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The mass of a planet that has a moon whose time period and orbital radius are T and R respectively can be written as -

$4 π^{2} R^{3} G^{‐ 1} T^{‐ 2}$

$8 π^{2} R^{3} G^{‐ 1} T^{‐ 2}$

$12 π^{2} R^{3} G^{‐ 1} T^{‐ 2}$

$16 π^{2} R^{3} G^{‐ 1} T^{‐ 2}$

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Explanation

(a) $m ω^{2} R = \frac{G M m}{R^{2}} \Rightarrow {(\frac{2 π}{T})}^{2} R = \frac{G M}{R^{2}} \Rightarrow M = \frac{4 π^{2} R^{3}}{G T^{2}}$

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The eccentricity of earth's orbit is 0.0167. The ratio of its maximum speed in its orbit to its minimum speed is

2.507

1.033

8.324

1.000

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Explanation

(b) $\frac{v_{m a x}}{v_{m i n}} = \frac{1 + e}{1 - e} = \frac{1 + 0.0167}{1 - 0.0167} = 1.033$

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The radius of orbit of a planet is two times that of the earth. The time period of planet is -

4.2 years

2.8 years

5.6 years

8.4 years

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Explanation

(b) $T_{2} = T_{1} {[\frac{R_{2}}{R_{1}}]}^{3 / 2} = 1 \times {(2)}^{3 / 2} = 2.8 y e a r s$

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The orbital angular momentum of a satellite revolving at a distance r from the centre is L. If the distance is increased to 16r, then the new angular momentum will be -

16 L

64 L

$\frac{L}{4}$

4 L

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Explanation

(d) $L = m v r = m \sqrt{\frac{G M}{r} r} = m \sqrt{G M r} ∴ L \propto \sqrt{r}$

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The condition for a uniform spherical mass m of radius r to be a black hole is [G= gravitational constant and g= acceleration due to gravity] .

${(2 G m / r)}^{1 / 2} \leq c$

${(2 G m / r)}^{1 / 2} = c$

${(2 G m / r)}^{1 / 2} \geq c$

${(g m / r)}^{1 / 2} \geq c$

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Explanation

$v_{e}$ should be more than or equal to speed of light.

i.e. $\sqrt{\frac{2 G m}{r}} \geq c$

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There are two bodies of masses 100 kg and 10000 kg separated by a distance 1 m. At what distance from the smaller body, the intensity of gravitational field will be zero ?

$\frac{1}{9} m$

$\frac{1}{10} m$

$\frac{1}{11} m$

$\frac{10}{11} m$

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Explanation

(c) $\frac{G \times 100}{x^{2}} = \frac{G \times 10000}{(1 - x^{2})} \Rightarrow \frac{10}{x} = \frac{100}{1 - x} \Rightarrow x = \frac{1}{11} m$

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What is the intensity of gravitational field of the centre of a spherical shell ?

$G m / r^{2}$

Zero

None of these

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Explanation

so, work done = 0

Hence, intensity of gravitational field of the centre of spherical shell is zero.

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The depth d at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface, is [R = radius of the earth]

$\frac{R}{n}$

$R [\frac{n - 1}{n}]$

$\frac{R}{n^{2}}$

$R [\frac{n}{n + 1}]$

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Explanation

(b) $g^{1} = g [1 - \frac{d}{R}] \Rightarrow \frac{g}{n} = g [1 - \frac{d}{R}] \Rightarrow d = [\frac{n - 1}{n}] R$

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The diameters of two planets are in the ratio 4 : 1 and their mean densities in the ratio 1 : 2. The acceleration due to gravity on the planets will be in ratio

1 : 2

2 : 3

2 : 1

4 : 1

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Explanation

(c) $g = \frac{4}{3} G πRp \Rightarrow \frac{g_{1}}{g_{2}} = \frac{p_{1} R_{1}}{p_{2} R_{2}} = \frac{1}{2} \times \frac{4}{1} = \frac{2}{1}$

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At what altitude in metre will the acceleration due to gravity be 25% of that at the earth's surface (Radius of earth = R metre)

$\frac{1}{4} R$

$\frac{3}{8} R$

$\frac{R}{2}$

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Explanation

b) $g' = g {[\frac{R}{R + h}]}^{2} \Rightarrow \frac{g}{4} = g {[\frac{R}{R + h}]}^{2} \Rightarrow \frac{1}{2} = \frac{R}{R + h} \Rightarrow R + h = 2 R S o, h = R$

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