The satellite which appears stationary above, observed from the earth’s surface is called geo-stationary satellite. The time period of such satellite is equal to that of the rotation of the earth, i. e, 24 hrs and the direction of revolution of the satellite the earth is same. Also, the center of the orbit of satellite and the earth’s center are coincident.
In a very massive stars, the gravity may become so strong that even light emitted from its surface does not escape. In other words, the escape velocity of such stars is greater than the speed of light. So, it is completely black. This is called black hole. The surface surrounding the black hole is called horizon. It encloses the collapsed matter completely. The horizon is an ideal one way membrane, i. e, particles and light can go inward through the surface but cannot go outward. As a result, the object is dark i. e, black and hides from view a finite region of space. At a point far from a black hole, its gravitation effects are the same as those of any normal body having same mass.
The radius of the black hole is given by,
$$R_H = \frac{2GM}{c^2}$$
where G is the universal Gravitation Constant, M is mass ofblack hole and c is speed of light in vacuum. RH is called Schwarzschild radius.
The force of gravitation attraction due to earth on a body depends on the mass m of the body i. e,
$$ F=\frac{GMm}{R^2} $$ (M and R is the mass and radius of earth)
Hence, heavier bodies are attracted more strongly bythe earth.But the acceleration due to gravityofthe falling body does not depend on the mass of the body as given by below equation,
$$g = \frac{GM}{R^2} $$
i.e., acceleration due to gravity is independent of themass of the body. Hence, all bodies fall with same time.
The variation of acceleration due to gravity with rotation of earth is given by, $$ g'=g-\omega^2 R cos^2 \phi $$
where is the latitude of any point at earth surface.
If the earth stops rotating about its axis, then
$$ \omega = 0 $$
$$ \therefore g' = g $$
This shows that the acceleration due to gravity will be equal to g at all parts around the surface of the earth.
The acceleration due to gravity on the surface of the earth is given by
$$ g = \frac{GM}{R^2} $$
when the radius decreases by 2% i.e., radius is
$$ = R - \frac{2}{100} \times R $$
$$= 0.98 R$$
The acceleration due to gravity becomes,
$$ g' = \frac{GM}{(0.98R)^2} $$
$$ = 1.04 \frac{GM}{R^2} $$
$$ = 1.04 g $$
$$ g' = 1.04 g $$
Thus, the percentage change in acceleration due to gravity is
$$ = \frac{g'-g}{g} \times 100 $$
$$ = \frac{1.04g-g}{g} \times 100 $$
$$ = 4% $$
Hence, the acceleration due to gravity increases by 4%
If the sun somehow collapsed to form black hole its radius decreases but the mass of the sun and the distance between the sun and the earth remains same. Thus, the force of gravitation between the sun and the earth remains same as given by the expression
$$ F = \frac{GM_sM_e}{R^2} $$
So, the earth's orbit would not be affected.
The escape velocity on the surface of a heavenly body is given by $$ V_{esp} = \sqrt{2gR} $$
Since, the acceleration due to gravity on the surface of the moon and the radius of the moon is small, the value of escape velocity on moon’s surface is also small (~ 2.5 kms-1). The rms velocity of the molecules of most of the gases is very high than the value of escape velocity on the moon. So, the molecules of such gases have escaped out of the moon and there is no atmosphere on moon.
When a space capsule revolves around the earth, the gravitational force of the earth acts on the capsule providing the necessary centripetal force. Also, it experiences a tendency to move tangentially in a straight line due to its inertia of direction. This outward tendency balances the gravitational force. Then, the apparent weight of the astronaut in the space capsule under such condition will be zero. Hence, the astronaut feels weightless ness.
The variation of acceleration due to gravity with the depth of the earth is given by, $$ g' = g(1-\frac{x}{R}) $$
where x is the depth from the surface.
The weight of a body is the product of mass and acceleration due to gravity. Thus, the variation of the weight with the depth of the earth can be obtained by multiplying equn (i) with mass of the body,
$$W' = mg' =mg(1-\frac{x}{R})$$
$$W' =W(1-\frac{x}{R})$$
Hence, the weight of the body decreases with the depth, i. e, the weight of the body is less inside the earth than on the surface.
When a spaceship revolves around the earth, the gravitational force of the earth acts on the capsule providing the necessary centripetal force. Also, it experiences a tendency to move tangentially in a straight line due to its inertia of direction. This outward tendency balances the gravitational force. Then, the apparent weight of the astronaut in the space capsule under such condition will be zero. Hence, the astronaut feels weightless ness. But an astronaut on the moon which is also orbiting around the earth experiences gravity of moon because of its high mass.
When the spoon is released from the satellite, it will continue to travel at the same velocity of satellite, along the orbital path, due to inertia.