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If there were a smaller gravitational effect, which of the following forces do you think would alter in some respect ?

Viscous forces

Archimedes uplift

Electrostatic force

None of the above

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Explanation

(b) As it depends on the weight of the body.

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The acceleration due to gravity near the surface of a planet of radius R and density d is proportional to

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

$d R^{2}$

$\frac{d}{r}$

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Explanation

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The weight of a body at the centre of the earth is -

Zero

Infinite

Same as on the surface of the earth

None of the above

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Explanation

(a)

$F = \frac{G M r}{R^{3}} H e r e, r = 0 s o, F = 0$

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A spherical planet far out in space has a mass $M_{0}$ and diameter $D_{0}$ . A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity which is equal to

$G M_{0} / {D_{0}}^{2}$

$4 m G M_{0} / {D_{0}}^{2}$

$4 G M_{0} / {D^{2}}_{0}_{}$

$G m M_{0} / {D^{2}}_{0}$

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Explanation

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If the radius of a planet is R and its density is $ρ$ , the escape velocity from its surface will be

$V_{e} \propto p R$

$V_{e} \propto R \sqrt{ρ}$

$V_{e} \propto \frac{\sqrt{p}}{R}$

$V_{e} \propto \frac{1}{\sqrt{p} R}$

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Explanation

(b) $v_{e} = R \sqrt{\frac{8}{3} G πρ} s o, v_{e} \propto R \sqrt{ρ}$

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If the distance between two masses is doubled, the gravitational attraction between them

Is doubled

Becomes four times

Is reduced to half

Is reduced to a quarter

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Explanation

(d) $F \propto \frac{1}{r^{2}}$ .

If r becomes double , then F reduces to $\frac{F}{4}$

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If the earth stops rotating, the value of ‘g’ at the equator will

Increase

Remain same

Decrease

None of the above

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Explanation

(a) g increases due to absence of the centrifugal force.

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If acceleration due to gravity on the surface of a planet is two times that on surface of earth and its radius is double that of earth. Then escape velocity from the surface of that planet in comparison to earth will be -

2 v_e

3 v_e

4 v_e

None of these

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Explanation

(a) v= $\sqrt{2 g R} .$ If acceleration due to gravity and radius of the planet, both are double that of earth then escape velocity will be two times. i.e. $v_{p} = 2 v_{e}$

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A body weight W newton at the surface of the earth. Its weight at a height equal to half the radius of the earth will be

$\frac{w}{2}$

$2 \frac{w}{3}$

$4 \frac{w}{9}$

$\frac{8 w}{27}$

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Explanation

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The escape velocity of a rocket launched from the surface of the earth

Does not depend on the mass of the rocket

Does not depend on the mass of the earth

Depends on the mass of the planet towards which it is moving

None of the above

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Explanation

(a)

$v_{e} = \sqrt{\frac{2 G M}{R}} \Rightarrow T h e e s c a p e v e l o c i t y o f a r o c k e t l a u n c h e d f r o m t h e s u r f a c e o f t h e e a r t h D o e s n o t d e p e n d o n t h e m a s s o f t h e r o c k e t .$

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