Gravitation

A ball is dropped from the top of a 100 m high tower on a planet . In the last 1/2 sec before hitting the ground….

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Q: A ball is dropped from the top of a 100 m high tower on a planet . In the last 1/2 sec before hitting the ground , it covers a distance of 19 m . Acceleration due to gravity (in m/s2) near the surface on that planet is …..

Click to See Solution :
Ans: a = 8 m/s2
Sol: Let the time taken by ball to travel 81 m is t sec

So time taken to travel 100 m is (t + 1/2) sec

$\displaystyle s = ut + \frac{1}{2} a t^2 $

$\displaystyle 81 = \frac{1}{2} a t^2 $

$\displaystyle t = 9 \sqrt{\frac{2}{a}} $ …(i)

$\displaystyle 100 = \frac{1}{2} a (t + \frac{1}{2})^2 $

$\displaystyle (t + \frac{1}{2}) = 10 \sqrt{\frac{2}{a}} $ …(ii)

$\displaystyle (9 \sqrt{\frac{2}{a}} + \frac{1}{2}) = 10 \sqrt{\frac{2}{a}} $

$\displaystyle \frac{1}{2} = \sqrt{\frac{2}{a}}$

a = 8 m/s2

 

Acceleration due to gravity on the earth’s surface at the poles is g and angular velocity of the earth about axis…

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Q: Acceleration due to gravity on the earth’s surface at the poles is g and angular velocity of the earth about axis passing through the pole is ω . An object is weighed at the equator and at a height h above the poles y using a spring balance . If weight are found to be same then h is (h << R , where R is the radius of earth )

(a) $\large \frac{\omega^2 R^2}{2 g}$

(b) $\large \frac{\omega^2 R^2}{ g}$

(c) $\large \frac{\omega^2 R^2}{4 g}$

(d) $\large \frac{\omega^2 R^2}{8 g}$

Click to See Solution :
Ans: (a)
Sol: Acceleration due to gravity at equator ,$\large g_e = g – \omega^2 R $

Acceleration due to gravity at a height h above the poles

$\large g_h = g(1 – \frac{2 h}{R})$

Given that , ge = gh

$\large g – \omega^2 R = g(1 – \frac{2 h}{R}) $

$\large h = \frac{\omega^2 R^2}{2 g}$

 

A body is moving in a low circular orbit bout a planet of mass M and radius R . The radius of the orbit can be taken ….

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Q: A body is moving in a low circular orbit bout a planet of mass M and radius R . The radius of the orbit can be taken to be R itself . Then ratio of the speed of this body in the orbit to the escape velocity from the planet is

(a) $\sqrt{2}$

(b) 2

(c) $\frac{1}{\sqrt{2}}$

(d) 1

Click to See Solution :
Ans: (c)
Sol: Orbital velocity $\large v_o = \sqrt{\frac{G M}{R}}$ …(i)

Escape velocity $\large v_e = \sqrt{\frac{2 G M}{R}}$ …(ii)

On dividing ,

$\large \frac{v_o}{v_e} = \sqrt{\frac{1}{\sqrt{2}}}$

 

A body A of mass m is moving in a circular orbit of radius R about a planet . Another body B of mass m/2 collides ….

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Q: A body A of mass m is moving in a circular orbit of radius R about a planet . Another body B of mass m/2 collides with A with a velocity which is half (v/2) the instantaneous velocity v of A . The collision is perfectly inelastic . Then the combined body

(a) escapes from the planet’s gravitational field

(b) starts moving in an elliptical orbit around the planet

(c) falls vertically downwards towards the planet

(d) continues to move in a circular orbit

Click to See Solution :
Ans: (b)

 

A box weighs 196 N on a spring balance at the north pole . Its weight recorded on the same balance if it is shifted ….

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Q: A box weighs 196 N on a spring balance at the north pole . Its weight recorded on the same balance if it is shifted to the equator is close to (Take g = 10 m/s2 at the north pole and radius of earth = 6400 km)

(a) 194.32 N

(b) 194.66 N

(c) 195.66 N

(d) 195.32 N

Click to See Solution :
Ans: (d)
Sol: Weight at north pole , mg = 196 N

At north pole , g = 10 m/s2

R = 6400 km

$\large g_{equator} = g – \omega^2 R$

$\large m g_{equator} = m g – m \omega^2 R$

$\large = 196 – \frac{196}{10} \times \frac{6400 \times 10^3 \times 4 \pi^2}{(24 \times 60 \times 60)^2} $

= 195.32 N