A uniform spherical planet is rotating about its axis. The speed of a point on its equator is v and the effective acceleration due to gravity on the equator is one third its value at the poles. Calculate the escape velocity for a particle at the pole of the planet. Give your answer in multiple of v.

Consider a spherical planet which is rotating about its axis such that the speed of a point on its equator is v and the effective acceleration due to gravity on the equator is 1/3 of its value at the poles. What is the escape velocity for a particle at the pole of this planet.

Consider a spherical planet which is rotating about its axis such that the speed of a point on its equator is 'v' and the effective acceleration due to gravity on the equator is 1/3 of its value at the poles. What is the escape velocity for the particle at the pole of this planet?

A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is V . Due to the rotation of planet about its axis the acceleration due to gravity g at equator is 1//2 of g at poles. The escape velocity of a particle on the planet in terms of V .

A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is V . Due to the rotation of planet about its axis the acceleration due to gravity g at equator is 1//2 of g at poles. The escape velocity of a particle on the planet in terms of V .

As we go from pole to the equator, the effective value of acceleration due to gravity decreases due to

A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is 7.5 kms^(-1) . Due to the rotation of the planet about its axis, the acceleration due to gravity g at equator is 1//2 of g at poles. What is the escape velocity ("in km s"^(-1)) of a particle on the planet from the pole of the planet?

A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is 7.5 kms^(-1) . Due to the rotation of the planet about its axis, the acceleration due to gravity g at equator is 1//2 of g at poles. What is the escape velocity ("in km s"^(-1)) of a particle on the planet from the pole of the planet?

Consider a spherical planet rotating about its axis. The velocity of a point at equator is v. The angular velocity of this planet is such that it makes apparent value of ‘g’ at the equator half of value of ‘g’ at the pole. The escape speed for a polar particle on the planet expressed as multiple of v is :