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If V, R, and g denote respectively the escape velocity from the surface of the earth, the radius of the earth, and acceleration due to gravity, then the correct equation is:

$V = \sqrt{g R}$

V= $\sqrt{\frac{4}{3} g R^{3}}$

V = R $\sqrt{g}$

V= $\sqrt{2 g R}$

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Explanation

$v = \sqrt{\frac{2 G M}{R}} \sin c e, g = \frac{G M}{R^{2}} \Rightarrow v = \sqrt{2 g R}$

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The depth at which the effective value of acceleration due to gravity is $\frac{g}{4}$ is

$\frac{3 R}{4}$

$\frac{R}{2}$

$\frac{R}{4}$

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Explanation

b) $g' = g (1 - \frac{d}{R}) \Rightarrow \frac{g}{4} = g (1 - \frac{d}{R}) \Rightarrow d = \frac{3 R}{4}$

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The value of ‘g’ at a particular point is $9.8 m / s^{2}$ . Suppose the earth suddenly shrinks uniformly to half its present size without losing any mass. The value of ‘g’ at the same point (assuming that the distance of the point from the centre of earth does not shrink) will now be

$4.9 m / s e c^{2}$

$3.1 m / s e c^{2}$

$9.8 m / s e c^{2}$

$19.6 m / s e c^{2}$

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Explanation

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The acceleration due to gravity on a planet is same as that on earth and its radius is four times that of earth. What will be the value of escape velocity on that planet if it is $V_{d}$ on earth -

$V_{d}$

2 $V_{d}$

$4 V_{d}$

$\frac{V_{d}}{2}$

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Explanation

$(b) v = \sqrt{2 g R} \Rightarrow \frac{v_{p}}{v_{e}} = \sqrt{\frac{g_{p}}{g_{e}} \times \frac{R_{e}}{R_{p}}} = \sqrt{1 \times 4} = 2 s o, v_{p} = 2 v_{e}$

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If the radius of a planet is four times that of earth and the value of g is the same for both, the escape velocity on the planet will be:

11.2 km/s

5.6 km/s

22.4 km/s

None

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Explanation

$v_{e} = \sqrt{\frac{2 G M}{R}} a n d g = \frac{G M}{R^{2}} \Rightarrow v_{e} = \sqrt{2 g R} (F o r e a r t h) v_{P} = \sqrt{2 \times g \times 4 R} = 2 v_{e} \Rightarrow v_{p} = 2 \times 11.2 k m / s = 22.4 k m / s$

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The earth (mass = $6 \times 10^{24} k g$ ) revolves round the sun with angular velocity $2 \times 10^{‐ 7} r a d / s$ in a circular orbit of radius $1.5 \times 10^{8} k m$ . The force exerted by the sun on the earth in Newtons, is

$18 \times 10^{25}$

Zero

$27 \times 10^{39}$

$36 \times 10^{21}$

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Explanation

(d) $m = 6 \times 10^{24} k g, ω = 2 \times 10^{‐ 7} r a d s, R = 1.5 \times 10^{11} m$

The force exerted by the sun on the earth

$F = m ω^{2} R$

By substituting the value we can get,

$F = 36 \times 10^{21} N$

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If the radius and acceleration due to gravity both are doubled, escape velocity of earth will become

11.2 km/s

22.4 km/s

5.6 km/s

44.8 km/s

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Explanation

$(b) v = \sqrt{2 g R} . I f g a n d R b o t h a r e d o u b l e d t h e n v w i l l b e c o m e s t w o t i m e s i . e . 11.2 \times 2 = 22.4 k m / s$

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The gravitational force between two point masses $m_{1}$ and $m_{2}$ at separation r is given by $F = k \frac{m_{1} m_{2}}{r^{2}}$

The constant k

Depends on system of units only

Depends on medium between masses only

Depends on both (a) and (b)

Is independent of both (a) and (b)

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Explanation

(a) k represents gravitational constant which depends only on the system of units.

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The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R, the radius of the planet would be

$\frac{1}{4} R$

$\frac{1}{2} R$

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Explanation

(d) $g = \frac{4}{3} π p G R \Rightarrow \frac{R_{p}}{R_{ε}} = (\frac{g_{p}}{g_{ε}}) (\frac{p_{ε}}{p_{p}}) = (1) \times (\frac{1}{2}) \Rightarrow R_{p} = \frac{R_{ε}}{2} = \frac{R}{2}$

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If R is the radius of the earth and g the acceleration due to gravity on the earth's surface, the mean density of the earth is

$4 π G / 3 g R$

$3 π R / 4 g G$

$3 g / 4 π R G$

$π R G / 12 G$

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Explanation

$g = \frac{4}{3} \frac{π R^{3} \times G ρ}{R^{2}} \Rightarrow ρ = \frac{3 g}{4 π R G}$

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