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Gravitation

Question

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Time period of pendulum, on a satellite orbiting the earth, is

$\frac{1}{π}$

zero

$π$

infinity

Solution

D.

infinity

On an artificial satellite orbiting the earth the acceleration is given by $\frac{GM}{R^{2}}$ towards the centre of the earth.

Now for a body of mass m on the satellite the graviational force due to earth is $\frac{G Mm}{R^{2}}$ towards the centre of the earth.

Let the reaction force on the surface of the satellite be N, then

$\frac{G Mm}{R^{2}} - N = m (\frac{G M}{R^{2}})$

⇒ N = 0

That is on the satellite there is a state of weightlessness or g = 0

T = 2 $π \sqrt{\frac{l}{g}}$ = ∞

Hence T = ∞

Some More Questions From Gravitation Chapter

A solid sphere is rotating in free space. If the radius of the sphere is increased keeping mass same which one of the following will not be affected?

Which of the following statements is correct?
(a) Centre of the mass of a body always coincides with the centre of gravity of the body.
(b) Centre of the mass of a body is the point at which the total gravitational torque on the body is zero
(c) A couple on a body produce both translational and rotational motion in a body.
(d) Mechanical advantage greater than one means that small effort can be used to lift a large load.

Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will:

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