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The gravitational potential energy of a body of mass ‘m’ at the earth’s surface $- m g R_{e}$ . Its gravitational potential energy at a height $R_{e}$ from the earth’s surface will be (Here $R_{e}$ is the radius of the earth)

$- 2 m g R_{e}$

$2 m g R_{e}$

$\frac{1}{2} m g R_{e}$

$- \frac{1}{2} m g R_{e}$

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Explanation

(d) $∆ U = U_{2} - U_{1} = \frac{m g h}{1 + \frac{h}{R_{e}}} = \frac{m g R_{e}}{1 + \frac{R_{e}}{R_{e}}} = \frac{m g R_{e}}{2}$

$\Rightarrow U_{2} + (m g R_{e}) = \frac{m g R_{e}}{2} \Rightarrow U_{2} = - \frac{1}{2} m g R_{e}$

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If the angular speed of the earth is doubled, the value of acceleration due to gravity (g) at the north pole

Doubles

Becomes half

Remains same

Becomes zero

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Explanation

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Escape velocity of a body of 1 kg mass on a planet is 100 m/sec. Gravitational Potential energy of the body at the planet is -

– 5000 J

– 1000 J

– 2400 J

5000 J

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Explanation

(a) $V_{e} = \sqrt{\frac{2 G M}{R}} = 100 \Rightarrow \frac{G M}{R} = 5000 P o t e n t i a l e n e r g y U = - \frac{G M m}{r} = - 5000 J$

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At the surface of a certain planet, acceleration due to gravity is one-quarter of that on earth. If a brass ball is transported to this planet, then which one of the following statements is not correct ?

The mass of the brass ball on this planet is a quarter of its mass as measured on earth

The weight of the brass ball on this planet is a quarter of the weight as measured on earth

The brass ball has the same mass on the other planet as on earth

The brass ball has the same volume on the other planet as on earth

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Explanation

(a) Mass of the ball always remain constant. It does not depend upon the acceleration due to gravity

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A body of mass m is placed on the earth’s surface. It is taken from the earth’s surface to a height 3R . The change in gravitational potential energy of the body is -

$\frac{2}{3} m g R$

$\frac{3}{4} m g R$

$\frac{m g R}{2}$

$\frac{m g R}{4}$

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Explanation

(b) $∆ U = \frac{m g h}{1 + \frac{h}{R}} = \frac{m g \times 3 R}{1 + \frac{3 R}{R}} = \frac{3}{4} m g R$

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Weight of 1 kg becomes 1/6 kg on moon. If radius of moon is $1.768 \times 10^{6} m$ , then the mass of moon will be -

$1.99 \times 10^{30} k g$

$7.56 \times 10^{22} k g$

$5.98 \times 10^{24} k g$

$7.65 \times 10^{22} k g$

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Explanation

(d) $g_{m} = \frac{G M_{m}}{{R_{m}}^{2}}$ and $g_{m} = \frac{g_{e}}{6} = \frac{9.8}{6} m / s^{2} = 1.63 m / s^{2}$

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A geo-stationary satellite is orbiting the earth at a height of 6 R above the surface of earth, R being the radius of earth. The time period of another satellite at a height of 2.5 R from the surface of earth is

10 hr

$(6 / \sqrt{2}) h r$

6 hr

$6 \sqrt{2} h r$

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Explanation

(d) Distances of the satellite from the centre are 7R and 3.5R respectively.

$\frac{T_{2}}{T_{1}} = {(\frac{R_{2}}{R_{1}})}^{\frac{3}{2}} \Rightarrow T_{2} = 24 {(\frac{3.5 R}{7 R})}^{\frac{3}{2}} = 6 \sqrt{2} h r$

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A body of mass m kg starts falling from a point 2R above the Earth’s surface. Its kinetic energy when it has fallen to a point ‘R’ above the Earth’s surface [R - Radius of Earth, M - Mass of Earth, G - Gravitational Constant] -

$\frac{1}{2} \frac{G M m}{R}$

$\frac{1}{6} \frac{G M m}{R}$

$\frac{2}{3} \frac{G M m}{R}$

$\frac{1}{3} \frac{G M m}{R}$

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Explanation

(b) Potential energy, U is given by

$U = - \frac{G M m}{R} = - \frac{G M m}{R + h} U_{i n i t i a l} = - \frac{G M m}{3 R} a n d U_{f i n a l} = - \frac{- G M m}{2 R} L o s s i n P E = g a i n i n K E = \frac{G M m}{2 R} - \frac{G M m}{3 R} = \frac{G M m}{6 R}$

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Let g be the acceleration due to gravity at earth's surface and K be the rotational kinetic energy of the earth. Suppose the earth's radius decreases by 2% keeping all other quantities same, then

g decreases by 2% and K decreases by 4%

g decreases by 4% and K increases by 2%

g increases by 4% and K increases by 4%

g decreases by 4% and K increases by 4%

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Explanation

c) $g = \frac{G M}{R^{2}}$ and $K = \frac{L^{2}}{2 I}$

If mass of the earth and its angular momentum remains constant then $g \propto \frac{1}{R^{2}}$ and $K \propto \frac{1}{R^{2}}$

i.e. if radius of earth decreases by 2% then g and K both increases by 4%.

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The distance between centre of the earth and moon is 384000 km. If the mass of the earth is $6 \times 10^{24} k g$ and $G = 6.66 \times 10^{- 11} N m^{2} / k g^{2}$ . The speed of the moon is nearly -

1 km/sec

4 km/sec

8 km/sec

11.2 km/sec

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Explanation

(a) $v = \sqrt{\frac{G M}{r}} = \sqrt{\frac{6.67 \times 10^{- 11} \times 6 \times 10^{24}}{384000 \times 10^{3}}} = 1 k m / s$

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