Gravitation and Orbital Mechanics

The satellite revolves around the Earth in a circular orbit when centrifugal force acting outward on the satellite is balanced by the gravitational force of the Earth. Thus, when gravitational force suddenly disappears, then only centrifugal force will be acting and velocity is tangential to the orbit and hence, the satellite will fly off tangentially. It is similar to an object tied with a string, fly off when string suddenly breaks.

Earth is about 147.1 million kilometres (91.4 million miles) from the Sun at perihelion around January 3, in contrast to about 152.1 million kilometres (94.5 million miles) at aphelion around July 4, a difference of about 5.0 million kilometres (3.1 million miles).

The satellite is revolving around earth because the centripetal force is balanced by earth's gravitational pull.If the gravitational pull disappears, the satellite free of centripetal force. So, it will travel with its instantaneous velocity i.e. in the direction tangential to the circular path.

If the satellite is brought to standstill $v=0$ which gives the net force on the satellite=
$F=\dfrac{GMm}{{a}^{2}} $ in the direction of the centre of the earth which earlier was providing centripetal acceleration. So the satellite will fall on the earth

Using Kepler's Third Law and the relationship between orbital periods and distances, the ratio of the masses can be derived from the given orbital data.

For any circular motion the angular momentum is conserved as no torque is acting on it because centripetal force acts through the point of axis.

$mr\omega^{2} = \dfrac {GMm}{r^{2}}$
$\Rightarrow r\omega^{2} = \dfrac {GM}{r^{2}}$
$\Rightarrow r^{3} = \dfrac {GM}{\omega^{2}} = \dfrac {GM}{R^{2}} . \dfrac {R^{2}}{\omega^{2}}$
$\Rightarrow r^{3} = g \dfrac {R^{2}}{\omega^{2}}$
Hence (D) is correct.

Kepler's third law states that the square of the orbital period is proportional to the cube of the semi-major axis (distance from the Sun). Therefore, planets further from the Sun have longer orbital periods. Mars (between Earth and Jupiter) must have a period between 1 and 11.8 years. Uranus (further than Neptune) would have a period longer than 165 years. The logic in option 2 and 4 fits the observed orbital distances.

• A single geostationary satellite is on a line of sight with about 40 percent of the earth's surface. 
  
• Three such satellites, each separated by 120 degrees of longitude, can provide coverage of the entire plane the north and south geographic poles.

Angular momentum of planet revolving around the sun is constant because the line of force of the gravitational force passes through the  sun . Therefore torque about the sun is zero.
$\therefore$  Angular momentum will remain constant.  And there is no other force acting on the planet, which results into zero tangential force.

Angular momentum is conserved if net external torque is zero
(A) torque about sun is zero , $\because$ line of focus passer though it and no external force is present.
(B) Torque of friction is present 
(C) Force on the electron is internal , $\therefore$ torque is zero 
(D) No external force is present which may produce changes in angular momentum.

The shape of the orbit a planet depends on angular momentum.

Gravity acts as a centripetal force, pulling objects inward toward the center of the larger body, which keeps them in a curved orbital path.

An orbiting satellite is in constant free fall; gravity pulls it toward Earth, while its forward velocity (inertia) keeps it from crashing, resulting in a stable orbit.