NCERT Solutions for Class 11 Physics Chapter 8

Q.1 Answer the following:
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
(b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?
(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (You can check this yourself using the data available in the succeeding exercises).
However, the tidal effect of the moon’s pull is greater than the tidal effect of sun. Why?

Ans.

(a) Gravitational influence of matter on nearby objects cannot be shielded by any means. This is due to the fact that gravitational force is independent of the nature of the material medium, however, it is not so in the case of the electrical forces. It is also independent of the presence of other objects.

(b)Yes, if the size of the space ship orbiting around the Earth is very large, then the astronaut can identify the change in Earth’s gravity (g).

(c) Tidal effect is inversely proportional to the cube of the distance. However, the gravitational force is inversely proportional to the square of the distance. The distance between the Moon and the Earth is very small in comparison to the distance between the Sun and the Earth. Therefore, the tidal effect of the Moon’s pull is more than the tidal effect of the Sun’s pull.

Q.2 Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?

Ans.

\begin{array}{l}\text{Since}\text{time taken by the Earth to revolve}\text{once around the}\text{Sun,}{\text{T}}_{\text{E}}\text{= 1 year}\\ {\text{Orbital radius of Earth, R}}_{\text{E}}\text{= 1 AU}\\ \text{Time taken by the given}\text{planet to revolve}\text{once around the}\text{Sun,}{\text{T}}_{\text{P}}\text{=}\frac{\text{1}}{\text{2}}{\text{T}}_{\text{E}}\text{= 0}\text{.5}\text{year}\\ \text{Let}\text{orbital radius of the}{\text{given planet = R}}_{\text{P}}\\ \text{According}\text{to Kepler’s third law of planetary motion, we have:}\\ {\text{R}}_{\text{P}}{\text{= R}}_{\text{E}}{\left(\frac{{\text{T}}_{\text{P}}}{{\text{T}}_{\text{E}}}\right)}^{\text{2/3}}\\ {\text{R}}_{\text{P}}\text{= 1}{\left(\frac{\text{0}\text{.5}\text{year}}{\text{1}\text{year}}\right)}^{\text{2/3}}\text{= 0}\text{.63}\text{AU}\\ \therefore \text{The orbital radius of the planet will be 0}\text{.63 times smaller than the orbital radius}\text{of}\text{Earth}\text{.}\end{array}

Q.3 Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 10⁸ m. Show that the mass of Jupiter is about one-thousandth that of the sun.

Ans.

\begin{array}{l}\text{Here,orbital period of}\text{a}\text{satellite}\text{of}\text{Jupiter,}{\text{T}}_{\text{1}}\text{= 1}\text{.769}\text{days = 1}\text{.769×24×60×60}\text{s}\\ \text{Orbital radius of}\text{satellite,}{\text{r}}_{\text{1}}\text{= 4}{\text{.22×10}}^{\text{8}}\text{m}\\ \text{G = Universal gravitational constant}\\ \text{Orbital radius of the Earth, r = 1}\text{A}\text{.U}\text{. = 1}{\text{.496×10}}^{\text{11}}\text{m}\\ \text{Mass of the Jupiter is given by,}{\text{M}}_{\text{J}}\text{=}\frac{{\text{4π}}^{\text{2}}{\text{r}}_{\text{1}}^{\text{3}}}{{\text{GT}}_{\text{1}}^{\text{2}}}\\ {\text{M}}_{\text{J}}\text{=}\frac{{\text{4π}}^{\text{2}}\text{×}{\left(\text{4}{\text{.22×10}}^{\text{8}}m\right)}^{\text{3}}}{\text{G×}{\left(\text{1}\text{.769×24×60×60 s}\right)}^{\text{2}}}\mathrm{\dots }\text{(i)}\\ \text{Now,}\text{orbital}\text{period}\text{of}\text{Earth}\text{around}\text{the}\text{Sun,}\text{T = 1}\text{year = 365}\text{.25×24×60×60}\text{s}\\ \text{Mass}\text{of}\text{Sun}\text{is}\text{given}\text{by,}{\text{M}}_{\text{S}}\text{=}\frac{{\text{4π}}^{\text{2}}{\text{r}}_{}^{\text{3}}}{{\text{GT}}_{}^{\text{2}}}\\ {\text{M}}_{\text{S}}\text{=}\frac{{\text{4π}}^{\text{2}}\text{×}{\left(\text{1}{\text{.496×10}}^{\text{11}}\text{m}\right)}^{\text{3}}}{\text{G×}{\left(\text{365}\text{.25×24×60×60 s}\right)}^{\text{2}}}\mathrm{\dots }\text{(ii)}\\ \text{Dividing}\text{equation}\text{(ii)}\text{by}\text{(i),}\text{we}\text{get}\\ \frac{{\text{M}}_{\text{S}}}{{\text{M}}_{\text{J}}}\text{=}\frac{{\text{4π}}^{\text{2}}\text{×}{\left(\text{1}{\text{.496×10}}^{\text{11}}\text{m}\right)}^{\text{3}}}{\text{G×}{\left(\text{365}\text{.25×24×60×60}\text{s}\right)}^{\text{2}}}\text{×}\frac{\text{G×}{\left(\text{1}\text{.769×24×60×60}\text{s}\right)}^{\text{2}}}{{\text{4π}}^{\text{2}}\text{×}{\left(\text{4}{\text{.22×10}}^{\text{8}}\text{m}\right)}^{\text{3}}}\text{= 1045}\text{.04}\\ \therefore \frac{{\text{M}}_{\text{S}}}{{\text{M}}_{\text{J}}}\approx \text{1000}\\ \therefore {\text{M}}_{\text{J}}\approx \frac{\text{1}}{\text{1000}}{\text{M}}_{\text{S}}\\ \therefore \text{It can be concluded that the mass of Jupiter}\text{is}\text{about one-thousandth that of the}\text{Sun}\text{.}\end{array}

Q.4 Let us assume that our galaxy consists of 2.5 × 10¹¹ stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be 10⁵ ly.

Ans.

\begin{array}{l}\text{Mass of our galaxy, M = 2}{\text{.5×10}}^{\text{11}}\text{solar mass}\\ \text{Mass of Sun = 2}{\text{.0 × 10}}^{\text{36}}\text{kg}\\ \therefore \text{Mass of our galaxy, M = 2}{\text{.5×10}}^{\text{11}}{\text{×2×10}}^{\text{36}}{\text{kg = 5×10}}^{\text{41}}\text{kg}\\ \text{Diameter of our}{\text{galaxy, d = 10}}^{\text{5}}\text{ly}\\ \text{Radius of our}{\text{galaxy, r = 5×10}}^{\text{4}}\text{ly}\\ \text{1 ly = 9}{\text{.46×10}}^{\text{15}}\text{m}\\ \therefore {\text{r = 5×10}}^{\text{4}}\text{×9}{\text{.46×10}}^{\text{15}}\text{m = 4}{\text{.73 ×10}}^{\text{20}}\text{m}\\ \text{As a star revolves around the galactic centre of our}\text{galaxy, its time period is,}\text{T =}\left(\frac{{\text{4π}}^{\text{2}}{\text{r}}_{}^{\text{3}}}{{\text{GT}}_{}^{\text{2}}}\right)\\ \text{T =}{\left(\frac{{\text{4π}}^{\text{2}}{\text{r}}_{}^{\text{3}}}{\text{GM}}\right)}^{1/2}\text{=}\left[\frac{\text{4×}{\left(\text{3}\text{.14}\right)}^{\text{2}}\text{×}{\left(\text{4}\text{.73}\right)}^{\text{3}}{\text{×10}}^{\text{60}}{\text{m}}^{\text{3}}}{\text{6}{\text{.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}{\text{×5×10}}^{\text{41}}\text{kg}}\right]\text{= 1}{\text{.12×10}}^{\text{16}}\text{s}\end{array}

Q.5 Choose the correct alternative:
(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
(b) The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence.

Ans.

(a) The total mechanical energy of a satellite is equal to the sum of its kinetic energy (always positive) and potential energy (maybe negative). The gravitational P.E. of the satellite is zero at infinity. The total energy of the satellite is negative as the Earth-satellite system is a bound system. Therefore, at infinity, the total energy of an orbiting satellite is equal to the negative of its kinetic energy.

(b) An orbiting satellite attains a definite amount of energy that enables it to revolve around the Earth. This energy is provided by the orbit of the satellite. It needs relatively lesser energy to move out of the effect of the Earth’s gravitational field than a stationary object on the Earth’s surface that initially contains no energy.

Q.6 Does the escape speed of a body from the earth depend on (a) the mass of the body, (b)the location from where it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched?

Ans.

$$\begin{array}{l}\text{The}\text{escape velocity of a body from the Earth is given}\\ \text{as:}{\text{v}}_{\text{esc}}=\sqrt{2gR}\end{array}$$

Here, g is the acceleration due to gravity and R is the radius of the Earth.

From the given relation, we observe that the escape velocity v_esc is independent of the mass of the body and the direction of its projection. But, it depends on the location from where the body is projected. As the gravitational potential depends slightly on the height, therefore, escape velocity depends slightly on the height of the location from where the body is launched.

Q.7 A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.

Ans.

For the given situation, the comet has constant values of angular momentum and total energy at all points of the orbit. But, its linear speed, angular speed, kinetic, and potential energy change with locations in the orbit.

Q.8 Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet (b) swollen face, (c) headache, (d) orientational problem?

Ans.

The symptoms (b), (c), and (d) can afflict an astronaut in space.

(a) Legs carry the weight of the body in a standing position because of gravitational pull. The astronaut in space is in weightlessness state because of the absence of gravity.

Thus, swollen feet do not affect an astronaut in space.

(b) A swollen face is caused because of apparent weightlessness in space. Since the sense organs such as eyes, ears nose, and mouth constitute a person’s face, hence, the swollen face can affect to great extent the seeing, hearing, smelling and eating capabilities of an astronaut in space.

(c) Headache is caused due to mental strain. The headache will have the same effect on the astronaut in space as on a person on earth, therefore, it can affect the working of an astronaut in space.

(d) Space has several orientations and frames of reference in space. Hence, the orientation problems can seriously affect an astronaut in space.

Q.9 In the following two exercises, choose the correct answer from among the given ones:
The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig 8.12) (i) a, (ii) b, (iii) c, (iv) 0.

Ans.

Since the gravitational potential (V) is constant everywhere inside a spherical shell, therefore, the gradient of gravitational potential

\left(\frac{dV}{dr}\right)

is zero everywhere inside the spherical shell. Since the gravitational intensity is the negative of the gradient of gravitational potential, therefore, the gravitational intensity is also zero everywhere inside the spherical shell. This shows that gravitational forces acting at a point inside a spherical shell are symmetric.

If we remove the upper half of a spherical shell, then the net gravitational force acting on a particle located at centre C will act downwards.

Since the gravitational intensity at a point is the gravitational force per unit mass at that point, it will also act downwards. Therefore, the gravitational intensity at the centre C will be along arrow c. Hence, the option (iii) is correct.

Q.10 Choose the correct answer from among the given ones:
For the problem 8.10, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.

Ans.

Since the gravitational potential (V) is constant everywhere inside a spherical shell, therefore, the gradient of gravitational potential

\left(\frac{dV}{dr}\right)

is zero everywhere inside the spherical shell. Since the gravitational intensity is equal to the negative of the gradient of gravitational potential, therefore, the gravitational intensity is also zero everywhere inside the spherical shell. This shows that gravitational forces acting at a point inside a spherical shell are symmetric.

If we remove the upper half of a spherical shell, then the net gravitational force acting on a particle located at centre C will act downwards.

Since the gravitational intensity at a point is the gravitational force per unit mass at that point, it will also act downwards. Therefore, the gravitational intensity at the centre P will be along arrow e.

Hence, (ii) is correct.

Q.11 A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the sun = 2 ×10³⁰ kg, mass of the earth = 6 × 10²⁴ kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 10¹¹ m).

Ans.

\begin{array}{l}\text{Here,}{\text{mass of sun, M}}_{\text{S}}{\text{= 2×10}}^{\text{30}}\text{kg}\\ {\text{Mass of earth, M}}_{\text{E}}{\text{= 6×10}}^{\text{24}}\text{kg}\\ \text{Orbital radius}\text{of}\text{Earth, r = 1}{\text{.5×10}}^{\text{11}}\text{m}\\ \text{Let}\text{mass of the rocket be m}\\ \text{If x is the distance from the centre of the Earth where the gravitational force}\text{acting on satellite Q becomes}\\ \text{equal}\text{to}\text{zero}\text{. Using Newton’s law of gravitation, we can equate gravitational forces acting on}\text{satellite Q}\\ \text{under the effect of the Sun and the Earth as:}\\ \frac{{\text{GmM}}_{\text{S}}}{{\left(\text{r-x}\right)}^{\text{2}}}\text{= G}\frac{{\text{mM}}_{\text{E}}}{{\text{x}}^{\text{2}}}\\ {\left(\frac{\text{r-x}}{\text{x}}\right)}^{\text{2}}\text{=}\frac{{\text{M}}_{\text{S}}}{{\text{M}}_{\text{E}}}\\ \therefore \frac{\text{r-x}}{\text{x}}\text{=}{\left(\frac{{\text{2 × 10}}^{\text{30}}\text{kg}}{{\text{6 × 10}}^{\text{24}}kg}\right)}^{\frac{\text{1}}{\text{2}}}\text{= 577}\text{.35}\\ \text{1}{\text{.5×10}}^{\text{11}}\text{m – x = 577}\text{.35x}\\ \text{578}\text{.35x = 1}{\text{.5×10}}^{\text{11}}\text{m}\\ \therefore \text{x = 2}{\text{.59×10}}^{\text{8}}\text{m}\end{array}

Q.12 How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 × 10⁸ km.

Ans.

\begin{array}{l}\text{Here,}\text{orbital radius of the Earth, r = 1}{\text{.5×10}}^{\text{11}}\text{m}\\ \text{Period}\text{of}\text{revolution}\text{of Earth around the Sun,}\text{T = 1}\text{year = 365}\text{.25 days = 365}\text{.25×24×60×60 s}\\ \text{Universal gravitational constant,}\text{G = 6}{\text{.67×10}}^{\text{–11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{–2}}\\ \text{Let}\text{mass}\text{of}\text{Sun be M and}\text{mass}\text{of}\text{Earth be m}\\ \text{Gravitational}\text{force}\text{acting}\text{on}\text{Earth}\text{due}\text{to}{\text{Sun, F}}_{\text{g}}\text{=}\frac{\text{GMm}}{{\text{r}}^{\text{2}}}\\ \text{Let}\text{angular}\text{velocity}\text{of}\text{Earth}\text{around}\text{Sun be ω}\\ \text{Centripetal}\text{force}\text{acting}\text{on}{\text{Earth, F}}_{\text{c}}{\text{= mrω}}^{\text{2}}\text{= mr}{\left(\frac{\text{2π}}{\text{T}}\right)}^{\text{2}}\text{= mr}\frac{{\text{4π}}^{\text{2}}}{{\text{T}}^{\text{2}}}\\ \text{Since}\text{gravitational}\text{pull}\text{of}\text{Sun}\text{on}\text{Earth}\text{provides}\text{the}\text{necessary}\text{centripetal}\text{force,}\\ \therefore \frac{\text{GMm}}{{\text{r}}^{\text{2}}}\text{= mr}\frac{{\text{4π}}^{\text{2}}}{{\text{T}}^{\text{2}}}\\ \therefore \text{M =}\frac{{\text{4π}}^{\text{2}}{\text{r}}^{\text{3}}}{{\text{GT}}^{\text{2}}}\\ \text{M =}\frac{\text{4×}{\left(\text{3}\text{.14}\right)}^{\text{2}}\text{×}{\left(\text{1}{\text{.5×10}}^{\text{11}}m\right)}^{\text{3}}}{\text{6}{\text{.67×10}}^{\text{–11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{–2}}\text{×}{\left(\text{365}\text{.25×24×60×60 s}\right)}^{\text{2}}}\text{= 2}{\text{.0×10}}^{\text{30}}\text{kg}\\ \therefore {\text{Mass of the Sun = 2×10}}^{\text{30}}\text{kg}\end{array}

Q.13 A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50×10⁸ km away from the sun?

Ans.

\begin{array}{l}\text{Here,}{\text{seperation of the earth from the sun, r}}_{\text{E}}\text{= 1}{\text{.5×10}}^{\text{8}}\text{km = 1}{\text{.5×10}}^{\text{11}}\text{m}\\ \text{Let}{\text{time period of the earth be T}}_{\text{E}}\\ {\text{Time period of saturn, T}}_{\text{s}}\text{= 29}{\text{.5 T}}_{\text{E}}\\ \text{Let}\text{distance}\text{of}\text{saturn}\text{from}\text{the}s{\text{un be r}}_{\text{S}}\\ \text{Using}\text{Kepler’s}\text{third}\text{law}\text{of}\text{planetary}\text{motion,}\text{we}\text{have}\\ \text{T =}{\left(\frac{{\text{4π}}^{\text{2}}{\text{r}}^{\text{3}}}{\text{GM}}\right)}^{\text{1/2}}\\ \text{In}\text{case}\text{of}s\text{aturn}\text{and}s\text{un,}\text{we}\text{have:}\\ \frac{{\text{r}}_{\text{S}}{}^{\text{3}}}{{\text{r}}_{\text{E}}{}^{\text{3}}}\text{=}\frac{{\text{T}}_{\text{S}}{}^{\text{2}}}{{\text{T}}_{\text{E}}{}^{\text{2}}}\\ \therefore {\text{r}}_{\text{S}}{\text{= r}}_{\text{E}}{\left(\frac{{\text{T}}_{\text{S}}}{{\text{T}}_{\text{E}}}\right)}^{\text{2/3}}\\ {\text{r}}_{\text{S}}\text{= 1}{\text{.5×10}}^{\text{11}}\text{m}\times {\left(\frac{\text{29}{\text{.5 T}}_{\text{E}}}{{\text{T}}_{\text{E}}}\right)}^{\text{2/3}}\\ {\text{r}}_{\text{S}}\text{= 1}{\text{.5×10}}^{\text{11}}\text{m}\times {\left(\text{29}\text{.5}\right)}^{\text{2/3}}\text{= 14}{\text{.32×10}}^{\text{11}}\text{m}\\ \therefore \text{Distance}\text{between}\text{saturn}\text{and}\text{sun = 14}{\text{.32×10}}^{\text{11}}\text{m}\end{array}

Q.14 n A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

Ans.

\begin{array}{l}\text{Here,}\text{weight of body, W = 63 N}\\ \text{Let}\text{radius}{\text{of Earth be R}}_{\text{E}}\\ \text{Acceleration due to gravity at height h from the Earth’s surface is given as:}\\ \text{g’ =}\frac{\text{g}}{{\left(\frac{\text{1+h}}{{\text{R}}_{\text{E}}}\right)}^{\text{2}}}\mathrm{\dots }\text{(i)}\\ \text{Here,}\text{g = Acceleration due to gravity on the}\text{surface}\text{of Earth}\\ \text{Here,}\text{height,}\text{h =}\frac{{\text{R}}_{\text{E}}}{\text{2}}\mathrm{\dots }\text{(ii)}\\ \text{Substituting}\text{the}\text{value}\text{of}\text{h}\text{from}\text{equation}\text{(ii)}\text{in}\text{equation}\text{(i),}\text{we}\text{obtain:}\\ \text{g’ =}\frac{\text{g}}{{\left(\frac{{\text{R}}_{\text{E}}\text{+h}}{{\text{R}}_{\text{E}}}\right)}^{\text{2}}}\text{=}\frac{{\text{gR}}_{\text{E}}{}^{\text{2}}}{{\left({\text{R}}_{\text{E}}\text{+}\frac{{\text{R}}_{\text{E}}}{\text{2}}\right)}^{\text{2}}}\\ \text{g’ =}\frac{\text{g}}{{\left(\text{1+}\frac{\text{1}}{\text{2}}\right)}^{\text{2}}}\text{=}\frac{\text{4}}{\text{9}}\text{g}\\ \text{Weight of a body of mass m at height h,}\text{W’ = mg’}\\ \text{W’ = m×}\frac{\text{4}}{\text{9}}\text{g =}\frac{\text{4}}{\text{9}}\text{W}\text{=}\frac{\text{4}}{\text{9}}\text{×63}\text{N = 28}\text{N}\end{array}

Q.15 Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface?

Ans.

\begin{array}{l}\text{Let}\text{mass}\text{of}\text{the}\text{body be m}\\ \text{Given,}\text{weight of the body}\text{at the surface}\text{of}\text{Earth, W = mg = 250 N}\\ \text{Here,}{\text{R}}_{\text{E}}\text{= Radius of the Earth}\\ \text{Acceleration due to gravity at depth d is given as:}\\ \text{g’ =}\left(\text{1 –}\frac{\text{d}}{{\text{R}}_{\text{E}}}\right)\text{g}\\ \text{The}\text{body is located at depth,}\text{d =}\frac{\text{1}}{\text{2}}{\text{R}}_{\text{E}}\\ \therefore \text{g’ =}\left(\text{1 –}\frac{{\text{R}}_{\text{E}}}{{\text{2×R}}_{\text{E}}}\right)\text{g =}\frac{\text{1}}{\text{2}}\text{g}\\ \text{Weight of the body at depth d}\text{is}\text{given}\text{as:}\\ \text{W’ = mg’ =}\frac{\text{1}}{\text{2}}\text{mg =}\frac{\text{1}}{\text{2}}\text{W}\\ \text{W’ =}\frac{\text{1}}{\text{2}}\text{×250}\text{N = 125}\text{N}\end{array}

Q.16 A rocket is fired vertically with a speed of 5 kms^–1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 10²⁴ kg; mean radius of the earth = 6.4 × 10⁶ m; G= 6.67 × 10^–11 Nm²kg^–2.

Ans.

\begin{array}{l}\text{Here,}{\text{velocity of the rocket, v = 5 kms}}^{\text{-1}}{\text{= 5×10}}^{\text{3}}{\text{ms}}^{\text{-1}}\\ \text{Mass of Earth,}{\text{M}}_{\text{E}}\text{= 6}{\text{.0×10}}^{\text{24}}\text{kg}\\ \text{Radius of Earth,}{\text{R}}_{\text{E}}\text{= 6}{\text{.4×10}}^{\text{6}}\text{m}\\ \text{Height attained by rocket mass, m be h}\\ \text{At the Earth’surface,}\\ {\text{Total energy of rocket, E}}_{\text{T}}\text{= Kinetic energy + Potential energy}\\ {\text{E}}_{\text{T}}\text{=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{+}\left(\frac{{\text{-GM}}_{\text{E}}\text{m}}{{\text{R}}_{\text{E}}}\right)\\ \text{At highest point h,}\text{velocity}\text{of}\text{rocket,}\text{v = 0}\\ \text{At highest point h,}\text{Potential}\text{energy}\text{of}\text{rocket = –}\frac{{\text{GM}}_{\text{E}}\text{m}}{{\text{R}}_{\text{E}}\text{+h}}\\ \text{Total energy of rocket}\text{=}\text{0 +}\left(\frac{{\text{-GM}}_{\text{E}}\text{m}}{{\text{R}}_{\text{E}}\text{+h}}\right)\text{= –}\frac{{\text{GM}}_{\text{E}}\text{m}}{{\text{R}}_{\text{E}}\text{+h}}\\ \text{According}\text{to the law of conservation of energy, we have}\\ \text{Total energy of rocket at the Earth’s surface = Total energy at height h}\\ \frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{+}\left(\frac{{\text{-GM}}_{\text{E}}\text{m}}{{\text{R}}_{\text{E}}}\right)\text{= –}\frac{{\text{GM}}_{\text{E}}\text{m}}{{\text{R}}_{\text{E}}\text{+h}}\\ \frac{\text{1}}{\text{2}}{\text{v}}^{\text{2}}{\text{= GM}}_{\text{E}}\left(\frac{\text{1}}{{\text{R}}_{\text{E}}}\text{–}\frac{\text{1}}{{\text{R}}_{\text{E}}\text{+h}}\right)\\ \frac{\text{1}}{\text{2}}{\text{v}}^{\text{2}}\text{=}\frac{{\text{gR}}_{\text{E}}\text{h}}{{\text{R}}_{\text{E}}\text{+h}}\\ \text{Here,}\text{g =}\frac{\text{GM}}{{\text{R}}_{\text{E}}{}^{\text{2}}}\text{=9}\text{.8}{\text{ms}}^{\text{-2}}\\ \therefore {\text{v}}^{\text{2}}\left({\text{R}}_{\text{E}}\text{+ h}\right){\text{= 2gR}}_{\text{E}}\text{h}\\ \text{h =}\frac{{\text{R}}_{\text{E}}{\text{v}}^{\text{2}}}{{\text{2gR}}_{\text{E}}{\text{-v}}^{\text{2}}}\\ \text{h =}\frac{\text{6}{\text{.4×10}}^{\text{6}}\text{m×}{\left({\text{5 × 10}}^{\text{3}}{\text{ms}}^{\text{-1}}\right)}^{\text{2}}}{\text{2×9}{\text{.8 ms}}^{\text{-2}}\text{×6}{\text{.4×10}}^{\text{6}}\text{m –}{\left({\text{5 × 10}}^{\text{3}}{\text{ms}}^{\text{-1}}\right)}^{\text{2}}}\text{= 1}{\text{.6×10}}^{\text{6}}\text{m}\\ \text{Height attained by the rocket w}\text{.r}\text{.t}\text{. the centre of the Earth, H}\text{=}{\text{R}}_{\text{E}}\text{+h}\\ \text{H = 6}{\text{.4×10}}^{\text{6}}\text{m +1}{\text{.6×10}}^{\text{6}}\text{m = 8}{\text{.0×10}}^{\text{6}}\text{m}\end{array}

Q.17 The escape speed of a projectile on the earth’s surface is 11.2 kms^–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

Ans.

\begin{array}{l}\text{Here,}\text{escape speed of the projectile}\text{from}\text{Earth,}{\text{v}}_{\text{esc}}\text{= 11}\text{.2}{\text{kms}}^{\text{-1}}\\ \text{Velocity of}{\text{projection of the projectile, v}}_{\text{P}}{\text{= 3v}}_{\text{esc}}\\ \text{Let}\text{mass of the projectile be m}\\ \text{Let}{\text{velocity of the projectile far away from the Earth be v}}_{\text{f}}\\ \therefore \text{Total energy of the projectile on the Earth =}\frac{\text{1}}{\text{2}}{\text{mv}}_{\text{P}}{}^{\text{2}}\text{–}\frac{\text{1}}{\text{2}}{\text{mv}}_{\text{esc}}{}^{\text{2}}\\ \text{Gravitational potential energy of the projectile far away from the Earth = 0}\\ \therefore \text{Total energy of the projectile far away from the Earth =}\frac{\text{1}}{\text{2}}{\text{mv}}_{\text{f}}{}^{\text{2}}\\ \text{According}\text{to}\text{the}\text{law}\text{of}\text{conservation}\text{of}\text{energy,}\text{we}\text{have}\\ \frac{\text{1}}{\text{2}}{\text{mv}}_{\text{P}}{}^{\text{2}}\text{–}\frac{\text{1}}{\text{2}}{\text{mv}}_{\text{esc}}{}^{\text{2}}\text{=}\frac{\text{1}}{\text{2}}{\text{mv}}_{\text{f}}{}^{\text{2}}\\ {\text{v}}_{\text{f}}\text{=}\sqrt{{\text{v}}_{\text{P}}{}^{\text{2}}{\text{– v}}_{\text{esc}}{}^{\text{2}}}\\ {\text{v}}_{\text{f}}\text{=}\sqrt{{\left({\text{3v}}_{\text{esc}}\right)}^{\text{2}}\text{–}{\left({\text{v}}_{\text{esc}}\right)}^{\text{2}}}\text{=}\sqrt{\text{8}}{\text{v}}_{\text{esc}}\text{= 31}\text{.68}{\text{kms}}^{\text{-1}}\end{array}

Q.18 A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0×10²⁴ kg; radius of the earth = 6.4×10⁶ m; G = 6.67×10^–11 Nm²kg^–2.

Ans.

\begin{array}{l}\text{Given,}{\text{height of satellite, h = 400 km = 4×10}}^{\text{5}}\text{m = 0}{\text{.4×10}}^{\text{6}}\text{m}\\ \text{Mass of Earth, M = 6}{\text{.0×10}}^{\text{24}}\text{kg}\\ \text{Mass of satellite, m = 200 kg}\\ \text{Radius of Earth, R = 6}{\text{.4×10}}^{\text{6}}\text{m}\\ \text{Universal gravitational constant, G = 6}{\text{.67×10}}^{\text{–11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{–2}}\\ {\text{Total energy of the satellite at height h, E}}_{\text{T}}\text{= PE + KE}\\ {\text{E}}_{\text{T}}\text{=}\left(\frac{\text{-GMm}}{\text{R+h}}\right)\text{+}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\\ \text{Since}\text{orbital velocity of the satellite, v =}\sqrt{\frac{\text{GM}}{\text{R+h}}}\\ \therefore \text{Total energy of}\text{the}\text{orbiting}\text{satellite}\text{at height, h =}\frac{\text{1}}{\text{2}}\text{m}\left(\frac{\text{GM}}{\text{R+h}}\right)\text{–}\frac{\text{GMm}}{\text{R+h}}\text{= –}\frac{\text{1}}{\text{2}}\left(\frac{\text{GMm}}{\text{R+h}}\right)\\ \text{This is also}\text{called}\text{as}\text{the}\text{bound energy of the satellite}\text{.}\\ \text{Energy required to move the satellite out of its orbit =}\frac{\text{1}}{\text{2}}\left(\frac{{\text{GM}}_{\text{e}}\text{m}}{{\text{R}}_{\text{e}}\text{+h}}\right)\\ \text{=}\frac{\text{1}}{\text{2}}\text{×}\frac{\text{6}{\text{.67 × 10}}^{\text{–11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{–2}}\text{×6}{\text{.0 × 10}}^{\text{24}}\text{kg×200 kg}}{\left(\text{6}{\text{.4 × 10}}^{\text{6}}\text{m+0}{\text{.4 ×10}}^{\text{6}}m\right)}\text{= 5}{\text{.9× 10}}^{\text{9}}\text{J}\end{array}

Q.19 Two stars each of one solar mass (= 2×10³⁰ kg) are approaching each other for a head on collision. When they are at a distance 10⁹ km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 10⁴ km. Assume the stars to remain undistorted until they collide. (Use the known value of G).

Ans.

\begin{array}{l}\text{Here,}{\text{mass of each star, M = 2 × 10}}^{\text{30}}\text{kg}\\ {\text{Radius of each star, R = 10}}^{\text{4}}{\text{km = 10}}^{\text{7}}\text{m}\\ \text{Initial}{\text{distance between the stars, r = 10}}^{\text{9}}{\text{km = 10}}^{\text{12}}\text{m}\\ \text{For negligible speeds, v = 0}\\ \therefore \text{Total initial}\text{energy of two stars}\text{=}\frac{\text{-GMM}}{\text{r}}\text{+}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{=}\frac{\text{-GMM}}{\text{r}}\text{+ 0}\mathrm{\dots }\text{(i)}\\ \text{In}\text{case the stars are about to collide:}\\ \text{Let}\text{velocity of the stars be v}\\ \text{Separation between the centers of the stars = 2R}\\ \text{Final kinetic energy of both stars}\text{=}\frac{\text{1}}{\text{2}}{\text{Mv}}^{\text{2}}\text{+}\frac{\text{1}}{\text{2}}{\text{Mv}}^{\text{2}}{\text{= Mv}}^{\text{2}}\\ \text{Final potential energy of both stars}\text{=}\frac{\text{-GMM}}{\text{2R}}\\ \text{Total final}{\text{energy of the two stars = Mv}}^{\text{2}}\text{–}\frac{\text{GMM}}{\text{2R}}\mathrm{\dots }\text{(ii)}\\ \text{According}\text{to the law of conservation of energy, we have}\\ {\text{Mv}}^{\text{2}}\text{–}\frac{\text{GMM}}{\text{2R}}\text{=}\frac{\text{-GMM}}{\text{r}}\\ {\text{v}}^{\text{2}}\text{=}\frac{\text{-GM}}{\text{r}}\text{+}\frac{\text{GM}}{\text{2R}}\\ {\text{v}}^{\text{2}}\text{= GM}\left(\text{–}\frac{\text{1}}{\text{r}}\text{+}\frac{\text{1}}{\text{2R}}\right)\\ {\text{v}}^{\text{2}}\text{= 6}{\text{.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}{\text{×2×10}}^{\text{30}}\text{kg×}\left[\text{–}\frac{\text{1}}{{\text{10}}^{\text{12}}\text{m}}\text{+}\frac{\text{1}}{{\text{2×10}}^{\text{7}}\text{m}}\right]\approx \text{6}{\text{.67×10}}^{\text{12}}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}\\ \text{v =}\sqrt{\text{6}{\text{.67×10}}^{\text{12}}{\text{m}}^{\text{2}}{\text{s}}^{\text{-2}}}\text{= 2}{\text{.58×10}}^{\text{6}}{\text{ms}}^{\text{-1}}\end{array}

Q.20 Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the midpoint of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?

Ans.

\begin{array}{l}\text{The situation is shown in the given figure}\text{.}\\ \text{Here,}\text{mass of each sphere, M = 100 kg}\\ \text{Distance between the spheres, r = 1 m}\\ \text{Here,}\text{we}\text{assumed}\text{S to}\text{be the midpoint between the spheres}\text{.}\\ \text{Since}\text{gravitational force applied by each sphere at}\text{the}\text{midpoint}\text{will be}\text{equal}\text{and opposite,}\\ \therefore \text{Gravitational}\text{force at midpoint}\text{S}\text{=}\text{0}\\ \text{Gravitational potential at}\text{midpoint}{\text{S, V}}_{\text{G}}\text{=}\frac{\text{-GM}}{\left(\frac{\text{r}}{\text{2}}\right)}\text{–}\frac{\text{GM}}{\left(\frac{\text{r}}{\text{2}}\right)}\text{= -4}\frac{\text{GM}}{\text{r}}\\ {\text{V}}_{\text{G}}\text{=}\frac{\text{4×6}{\text{.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}\text{×100 kg}}{\text{1 m}}\text{= -2}{\text{.67×10}}^{\text{-8}}{\text{Jkg}}^{\text{-1}}\\ \text{Since effective force acting on an object placed at the mid-point is zero, therefore, any object placed}\\ \text{at point S will be in equilibrium}\text{. However, the equilibrium is unstable, because, any variation in the}\\ \text{position of the object will change the effective force in that direction}\text{.}\end{array}

Q.21 As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0 × 10²⁴ kg, radius = 6400 km.

Ans.

\begin{array}{l}\text{Here,}\text{mass of the Earth, M = 6}{\text{.0×10}}^{\text{24}}\text{kg}\\ \text{Radius of the Earth, R = 6400 km = 6}{\text{.4×10}}^{\text{6}}\text{m}\\ \text{Height of the}\text{geostationary satellite from the surface of Earth,}\text{h = 36000 km = 3}{\text{.6×10}}^{\text{7}}\text{m}\\ \text{Universal}\text{Gravitational}\text{constant,}\text{G}\text{=}\text{6}{\text{.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}\\ \text{Gravitational potential energy at}\text{height}\text{h}\text{due to Earth’s gravity}\text{is}\text{given}\text{as:}\\ \text{V =}\frac{\text{-GM}}{\left(\text{R+h}\right)}\\ \text{V =}\frac{\text{-6}{\text{.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}\text{×6}{\text{.0 × 10}}^{\text{24}}\text{kg}}{\text{3}{\text{.6×10}}^{\text{7}}\text{m + 0}{\text{.64×10}}^{\text{7}}\text{m}}\text{= -9}{\text{.4× 10}}^{\text{6}}{\text{Jkg}}^{\text{-1}}\end{array}

Q.22 A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (mass of the sun = 2×10³⁰ kg)

Ans.

\begin{array}{l}\text{The body will stuck to the surface of a star if the inward gravitational force is more}\text{than the outward}\\ \text{centrifugal}\text{force caused by the rotation of the star}\text{.}\\ \text{Here,}\text{radius of the star,}\text{R = 12 km = 1}{\text{.2 ×10}}^{\text{4}}\text{m}\\ \text{Mass of the star,}\text{M = 2}{\text{.5×2×10}}^{\text{30}}{\text{= 5×10}}^{\text{30}}\text{kg}\\ \text{Let}\text{mass of the body}\text{=}\text{m}\\ {\text{Gravitational force, f}}_{\text{g}}\text{=}\frac{\text{GMm}}{{\text{R}}^{\text{2}}}\\ \text{Here,}\text{G = 6}{\text{.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}\\ \therefore {\text{f}}_{\text{g}}\text{=}\frac{\text{6}{\text{.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}{\text{×5 × 10}}^{\text{30}}\text{kg×m}}{{\left(\text{1}{\text{.2 ×10}}^{\text{4}}\text{m}\right)}^{\text{2}}}\text{= 2}{\text{.31×10}}^{\text{11}}\text{mN}\\ {\text{Centrifugal force, f}}_{\text{c}}{\text{= mrω}}^{\text{2}}\\ \text{Angular speed, ω = 2πν}\\ {\text{f}}_{\text{c}}\text{= mR}{\left(\text{2πν}\right)}^{\text{2}}\\ \text{Here,}\text{angular frequency,}\text{ν = 1}{\text{.2 revs}}^{\text{–1}}\\ \therefore {\text{f}}_{\text{c}}\text{= m×(1}{\text{.2 ×10}}^{\text{4}}\text{m)×4×}{\left(\text{3}\text{.14}\right)}^{\text{2}}\text{×}{\left(\text{1}{\text{.2 revs}}^{\text{–1}}\right)}^{\text{2}}\text{= 1}{\text{.7×10}}^{\text{5}}\text{mN}\\ {\text{As f}}_{\text{g}}{\text{> f}}_{\text{c}}\text{, the body will remain stuck to the star}\text{.}\end{array}

Q.23

A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the Sun = 2 × 10³⁰ kg; mass of mars = 6.4 × 10²³ kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 10⁸ kg; G = 6.67 × 10^–11 Nm²kg^–2.

Ans.

\begin{array}{l}\text{Here,}{\text{mass of spaceship, m}}_{\text{S}}\text{= 1000 kg}\\ {\text{Mass of Sun, M = 2×10}}^{\text{30}}\text{kg}\\ {\text{Mass of Mars, m}}_{\text{M}}\text{= 6}{\text{.4×10}}^{\text{23}}\text{kg}\\ \text{Radius}\text{of}\text{the}\text{orbit of Mars, R = 2}{\text{.28×10}}^{\text{8}}\text{km = 2}{\text{.28 ×10}}^{\text{11}}\text{m}\\ \text{Radius of Mars, r = 3395 km = 3}{\text{.395×10}}^{\text{6}}\text{m}\\ \text{Universal gravitational constant, G = 6}{\text{.67×10}}^{\text{–11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{–2}}\\ \text{P}\text{.E}\text{. of the spaceship because}\text{of}\text{the gravitational}{\text{attraction of the Sun, (PE)}}_{\text{S}}\text{=}\frac{{\text{-GMm}}_{\text{S}}}{\text{R}}\\ \text{Potential energy of the spaceship because}\text{of the gravitational attraction of Mars,}{\text{(PE)}}_{\text{M}}\text{=}\frac{{\text{-Gm}}_{\text{M}}{\text{m}}_{\text{S}}}{\text{r}}\\ \text{As the spaceship is stationed on Mars, its velocity =0}\\ \therefore \text{Kinetic energy}\text{of the spaceship = 0}\\ {\text{Total energy of the spaceship, E}}_{\text{T}}\text{=}{\text{(PE)}}_{\text{S}}{\text{+ (PE)}}_{\text{M}}\\ {\text{E}}_{\text{T}}\text{=}\frac{{\text{-GMm}}_{\text{S}}}{\text{R}}\text{–}\frac{{\text{-Gm}}_{\text{M}}{\text{m}}_{\text{S}}}{\text{r}}{\text{= -Gm}}_{\text{S}}\left(\frac{\text{M}}{\text{R}}\text{+}\frac{{\text{m}}_{\text{M}}}{\text{r}}\right)\\ \text{The negative sign shows that the system is in bound state}\text{.}\\ \text{Energy required for launching the spaceship from the solar system = –}\left(\text{Total energy of the space}\text{ship}\right)\\ \text{= –}\left[{\text{Gm}}_{\text{S}}\left(\frac{\text{M}}{\text{R}}\text{+}\frac{{\text{m}}_{\text{M}}}{\text{r}}\right)\right]\\ {\text{= Gm}}_{\text{S}}\left(\frac{\text{M}}{\text{R}}\text{+}\frac{{\text{m}}_{\text{S}}}{\text{r}}\right)\\ \text{= 6}{\text{.67×10}}^{\text{–11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{–2}}\text{×1000 kg×}\left[\frac{{\text{2 × 10}}^{\text{30}}\text{kg}}{\text{2}{\text{.28 ×10}}^{\text{11}}\text{m}}\text{+}\frac{\text{6}{\text{.4 × 10}}^{\text{23}}\text{kg}}{\text{3}{\text{.395 × 10}}^{\text{6}}\text{m}}\right]\\ \text{= 5}{\text{.98×10}}^{\text{11}}\text{J}\end{array}

Q.24 A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 kms^–1. If 20% of its initial energy is lost due to Martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4×10²³ kg; radius of mars = 3395 km; G = 6.67× 10^-11 Nm²kg².

Ans.

\begin{array}{l}\text{Here,}\text{mass of the}\text{Mars, M = 6}{\text{.4×10}}^{\text{23}}\text{kg}\\ {\text{Initial velocity of the rocket, v = 2 kms}}^{\text{-1}}{\text{= 2×10}}^{\text{3}}{\text{ms}}^{\text{-1}}\\ \text{Radius of}\text{the Mars, R = 3395 km = 3}{\text{.395×10}}^{\text{6}}\text{m}\\ \text{Universal gravitational constant, G = 6}{\text{.67×10}}^{\text{–11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{–2}}\\ \text{Mass of the rocket be m}\\ \text{Initial K}\text{.E}\text{. of the rocket =}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\\ \text{Initial P}\text{.E}\text{. of the rocket}\text{=}\frac{\text{-GMm}}{\text{R}}\\ \text{Total initial energy}\text{of}\text{the}\text{rocket}\text{=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{–}\frac{\text{GMm}}{\text{R}}\\ \text{If 20 \% of initial kinetic energy is lost because}\text{of Martian atmospheric resistance,}\\ \text{then}\text{only 80 \% of its kinetic energy helps in reaching a height}\text{.}\\ \therefore \text{Total}\text{amount}\text{of initial energy available =}\frac{\text{80}}{\text{100}}\text{×}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\text{–}\frac{\text{GMm}}{\text{R}}\text{= 0}\text{.4}{\text{mv}}^{\text{2}}\text{–}\frac{\text{GMm}}{\text{R}}\\ \text{Let}\text{maximum height achieved by the rocket be h}\\ \text{At maximum height, the velocity of the rocket = 0}\\ \therefore \text{At maximum height, the K}\text{.E}\text{. of the rocket}\text{=}\text{0}\\ \text{At height h,}\text{total energy of the rocket = –}\frac{\text{GMm}}{\left(\text{R+h}\right)}\\ \text{According}\text{to the law of conservation of energy for the rocket, we have:}\\ \text{0}\text{.4}{\text{v}}^{\text{2}}\text{=}\frac{\text{GM}}{\text{R}}\text{–}\frac{\text{GM}}{\left(\text{R+h}\right)}\\ \frac{\text{R+h}}{\text{h}}\text{=}\frac{\text{GM}}{\text{0}\text{.4}{\text{v}}^{\text{2}}\text{R}}\\ \text{h =}\frac{\text{0}\text{.4}{\text{R}}^{\text{2}}{\text{v}}^{\text{2}}}{\text{GM-0}\text{.4}{\text{v}}^{\text{2}}\text{R}}\\ \text{h =}\frac{\text{0}\text{.4×}{\left(\text{3}{\text{.395 × 10}}^{\text{6}}\text{m}\right)}^{\text{2}}\text{×}{\left({\text{2× 10}}^{\text{3}}{\text{ms}}^{\text{-1}}\right)}^{\text{2}}}{\text{6}{\text{.67×10}}^{\text{–11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{–2}}\text{×6}{\text{.4 × 10}}^{\text{23}}\text{kg -0}\text{.4×}{\left({\text{2× 10}}^{\text{3}}{\text{ms}}^{\text{-1}}\right)}^{\text{2}}\text{×}\left(\text{3}{\text{.395 × 10}}^{\text{6}}\text{m}\right)}\text{= 495}\text{km}\end{array}