Deduce an expression for the minimum velocity with which a rocket must be fired to escape earth's gravitational field.

The velocity with which a projectile must be fired to escape from the earth does not depend upon

A particle of mass m is lying at the centre of a solid sphere of mass M and radius R . There is a turnel of negligible thickness, so that particle may escape. Find the minimum velocity required to escape the particle from the gravitational field of the sphere.

Find the escape speed from the earth for a 6000kg spacecraft and find the kinetic energy it must have at the surface of the earth in order to escape the Earth's gravitational field? Mass of the earth is 5.98 xx 10^(24)kg and its radius is 6.37 xx 10^(6) m

A satellite is revolving in a circular orbit at a height 'h' from the earth's surface (radius of earth R). The minimum increase in its orbital velocity required, So that the satellite could escape from the earth's gravitational field, is close to :(Neglect the effect of atomsphere.)

Assume that there is a smooth tunnel of depth (R )/(2) along a diameter of earth. A particle is projected from the bottom of tunnel with speed u. Find the minimum value of u so that the particle is able to escape the gravitational field of earth. M and R represents mass and radius of earth

Derive an expression for the escape velocity of a body from any planet.

If the earth has mass 9 times and radius twice that of the planet Mars, calculate the minimum velocity required by a rocket to pull out of the gravitational force of Mars. Escape velocity on the surface of the earth in 11.2 km//s

A satellite is revolving in a circular orbit at a height h from the earth surface ,such that hlt lt R is the readius of the earth .Assuming that the effect of earth 's atmosphere can be neglected the minimum increase in the speed required so that the stallite could escape from the gravitational field of earth is :

A rocket starts vertically upward with speed v_(0) . Show that its speed v at height h is given by v_(0)^(2)-v^(2)=(2hg)/(1+h/R) where R is the radius of the earth and g is acceleration due to gravity at earth's suface. Deduce an expression for maximum height reachhed by a rocket fired with speed 0.9 times the escape velocity.