The earth is moving on an elliptical path, whose one focus sun is situated as shown in figure. If `AS = r_(min)` and `SB = r_(max)`, the sun and earth system obey the Kepler's law, the square of time-period is directly proportional to

The earth E moves in an elliptical orbit with the sun S at one of the foci as shown in figure. Its speed of motion will be maximum at the point

The Earth E moves in an elliptical orbit with the Sun S at one of the foci as shown in figure. Its speed of motion will be maximum at a point .............

Assertion : The time period of a satellite revolving very close to the surface of earth is less. Reason : According to Kepler's law, square of time period of revolution is directly proportional to the cube of semi-major axis.

A planet of mass m is moving in an elliptical orbit about the sun (mass of sun = M). The maximum and minimum distances of the planet from the sun are r_(1) and r_(2) respectively. The period of revolution of the planet wil be proportional to :

A planet of mass m is moving in an elliptical orbit about the sun (mass of sun = M). The maximum and minimum distances of the planet from the sun are r_(1) and r_(2) respectively. The period of revolution of the planet wil be proportional to :

A planet of mass m is moving in an elliptical orbit about the sun (mass of sun = M). The maximum and minimum distances of the planet from the sun are r_(1) and r_(2) respectively. The period of revolution of the planet wil be proportional to :

If a and b are the nearest and farthest distances of a planet from the sun and the planet is revolving in an elliptical orbit, then square of the time period of revolution of that planets is directly proportional to

A planet is moving in an elliptical path around the sun as shown in figure. Speed of planet in positions P and Q are V_(1) and V_(2) respectively with SP=r_(1) and SQ=r_(2) , then v_(1)//v_(2) is equal to