Escape velocity from the surface of the earth is `V_e`. if the radius of the earth shrinks to half of the present value without any change in its mass, then escape velocity from the surface of the earth would be ?

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 from the surface of the earth is equal to

The value of escape speed from the surface of earth is

The escape velocity on the surface of the earth is 11.2 km/s. If mass and radius of a planet is 4 and 2 times respectively than that of earth, what is the escape velocity from the planet ?

The escape velocity of a particle of a particle from the surface of the earth is given by

An artificial satellite is moving in circular orbit around the earth with speed equal to half the magnitude of escape velocity from the surface of earth. R is the radius of earth and g is acceleration due to gravity at the surface of earth (R = 6400km) Then the distance of satelite from the surface of earth is .

A particle is throws vertically upwards from the surface of earth and it reaches to a maximum height equal to the radius of earth. The radio of the velocity of projection to the escape velocity on the surface of 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:

Mass and radius of a planet are two times the value of earth. What is the value of escape velocity from the surface of this planet?

A planet in a distant solar systyem is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is 11kms^(-1) , the escape velocity from the surface of the planet would be