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A spring balance is graduated on sea level. If a body is weighed with this balance at consecutively increasing heights from earth's surface, the weight indicated by the balance

Will go on increasing continuously

Will go on decreasing continuously

Will remain same

Will first increase and then decrease

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Explanation

(b) Because value of g decreases with increasing height

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The gravitational field due to a mass distribution is $E = K / x^{3}$ in the x-direction. (K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x is

K/x

K/2x

$K / x^{2}$

K/2 $x^{2}$

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Explanation

(d) Gravitational potential $= \int I d x = \int_{x}^{\infty} \frac{K}{x^{3}} d x = K {(\frac{x^{- 3 + 1}}{- 3 + 1})}_{x}^{\infty} = |\frac{- K}{2 x^{2}}| = \frac{K}{2 x^{2}}$

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The change in potential energy, when a body of mass m is raised to a height nR from the earth's surface is (R = Radius of earth)

$m g R \frac{n}{n - 1}$

nmgR

C.
mgR $\frac{n^{2}}{n^{2} + 1}$
D.

$m g R \frac{n}{n + 1}$

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Explanation

(d) $∆ U = \frac{m g h}{1 + \frac{h}{R}} = \frac{m g n R}{1 + \frac{n R}{R}} = \frac{n m g R}{n + 1}$

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If the earth suddenly shrinks (without changing mass) to half of its present radius, the acceleration due to gravity will be

g/2

g/4

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Explanation

(b) $g = \frac{G M}{R^{2}}$ . If radius shrinks to half of its present value then g will becomes four time

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The masses and radii of the earth and moon are $M_{1,} R_{1}$ and $M_{2}, R_{2}$ respectively. Their centres are distance d apart. The minimum velocity with which a particle of mass m should be projected from a point midway between their centres so that it escapes to infinity is

$2 \sqrt{\frac{G}{D} (M_{1} + M_{2})}$

$2 \sqrt{\frac{2 G}{D} (M_{1} + M_{2})}$

$2 \sqrt{\frac{G M}{D} (M_{1} + M_{2})}$

$2 \sqrt{\frac{G M (M_{1} + M_{2})}{D (R_{1} + R_{2})}}$

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Explanation

(a) Gravitational potential at mid point

$V = \frac{- G M_{1}}{d / 2} + \frac{- G M_{2}}{d / 2} N o w, P E = m \times V = \frac{- 2 G M}{d} (M_{1} + M_{2}) [m = m a s s o f p a r t i c l e] S o, f o r p r o j e c t i n g p a r t i c l e f r o m m i d p o i n t t o i n f i n i t y$

$K E = |P E| \Rightarrow \frac{1}{2} m v^{2} = \frac{2 G m}{d} (M_{1} + M_{2}) \Rightarrow v = 2 \sqrt{\frac{G (M_{1} + M_{2})}{d}}$

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If mass of earth is M, radius is R and gravitational constant is G, then work done to take 1 kg mass from earth surface to infinity will be

$\sqrt{\frac{G M}{2 R}}$

$\frac{G M}{R}$

$\sqrt{\frac{2 G M}{R}}$

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

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Explanation

(b) Potential energy of the 1 kg mass which is placed at the earth surface= $- \frac{G M}{R}$

its potential energy at infinite = 0

Work done = change in potential energy = $\frac{G M}{R}$

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If $v_{e}$ and $v_{o}$ represent the escape velocity and orbital velocity of a satellite corresponding to a circular orbit of radius R, then

$v_{e} = v_{o}$

$\sqrt{2} v_{o} = v_{e}$

$v_{e} = v_{o} / \sqrt{2}$

$v_{e}$ and $v_{o}$ are not related

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Explanation

(b) $v_{e} = \sqrt{2 g R} a n d v_{o} = \sqrt{g R} ∴ \sqrt{2} v_{o} = v_{e}$

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If r represents the radius of the orbit of a satellite of mass m moving around a planet of mass M, the velocity of the satellite is given by:

$v^{2} = g \frac{M}{r}$

$v^{2} = \frac{G M m}{r}$

$v = \frac{G M}{r}$

$v^{2} = \frac{G M}{r}$

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Explanation

$F_{e} = \frac{m v^{2}}{r} a n d F_{g} = \frac{G M m}{r^{2}}$

For the satellite to revolve in the orbit the centripetol force must be blamed by gravitational force i.e.

$F_{C} = F_{g} \Rightarrow \frac{m v^{2}}{r} = \frac{G M m}{r^{2}} \Rightarrow v^{2} = \frac{G M}{r}$

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An earth satellite of mass m revolves in a circular orbit at a height h from the surface of the earth. R is the radius of the earth and g is acceleration due to gravity at the surface of the earth. The velocity of the satellite in the orbit is given by

$\frac{g R^{2}}{R + h}$

$\frac{g R}{R + h}$

$\sqrt{\frac{g R^{2}}{R + h}}$

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Explanation

(d) $v_{o} = \sqrt{\frac{G M}{r}} = \sqrt{\frac{g R^{2}}{R + h}}$

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A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of earth in new orbit is 2 times that of the earlier orbit. The time period in the second orbit is

4.8 hours

$48 \sqrt{2}$ hours

24 hours

$24 \sqrt{2}$ hours

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Explanation

(b) $T \propto r^{3 / 2}$ . If r becomes double then time period will becomes ${(2)}^{\frac{3}{2}}$ times.
So new time period will be $24 \times 2 \sqrt{2} h r$ i.e. $T = 48 \sqrt{2}$

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