The escape velocity of a body from the earth is `V_(e)` if the radius of earth contracts to `1//4` th of its value keeping the mass of the earth constant the escape velocity will be

The escape velocity of a body from the earth's surface, v_("esc")= ……….. .

The escape velocity from the surface of the earth is (where R_(E) is the radius of the earth )

The escape velocity from the surface of the earth is (where R_(E) is the radius of the earth )

The escape velocity of a body on the surface of the Earth is 11.2km/s If the mass of the Earth is increase to twice its present value and the radius of the earth becomes half, the escape velocity becomes = ……"kms"^(-1)

The escape velocity of a body on the surface of earth is 11.2 km/ s. If the earth's mass increases to twice its present value and the radius of the earth becomes half, the escape velocity would become

The value of the escape velocity from the earth is v_e . If the radius of a planet is 4 times that of the earth and its density is 9 times the density of the earth, then the value of the escape velocity from that planet will be

The escape velocity of a body from the earth is V_(e) . The escape velocity of a planet whose mass and radius are twice those of the earth is

The escape velocity of a body from earth is about 11.2 km/s. Assuming the mass and radius of the earth to be about 81 and 4 times the mass and radius of the moon, the escape velocity in km/s from the surface of the moon will be: