## Question

A body is at a height equal to the radius of the earth from the surface of the earth. With what velocity be it thrown so that it goes out of the gravitational field of the earth? Given:

N m^{2} kg^{–2}.

### Solution

Escape velocity of a body from the earth’s surface is

Where *R _{e}* is the radius of the earth (distance of the body from the centre of the earth). If the body is at a height

*R*from the earth’s surface, then the distance of the body from the centre of the earth will be 2

_{e}*R*. Hence in this case, the escape velocity of the body will be

_{e}

Substituting the given values:

#### SIMILAR QUESTIONS

(i) A satellite is revolving in an orbit close to the earth’s surface. Taking the radius of the earth as find the value of the orbital speed and the period of revolution of the satellite.

(ii) What is the relationship of this orbital speed to the velocity required to send a body from the earth’s surface into space, never to return?

An artificial satellite revolving coplanar with the equator around the earth, appears stationary to an observer on the earth. Calculate the height of the satellite above the earth.

An artificial satellite is revolving at a height of 500 km above the earth’s surface in a circular orbit, completing one revolution in 98 minutes. Calculate the mass of the earth. Given:

If the period of revolution of an artificial satellite just above the earth be *T*and the density of earth be then prove that ρT^{2} is a universal constant. Also calculate the value of this constant.

A space-craft is launched in a circular orbit near the earth. How much more velocity will be given to the space-craft so that it will go beyond the attraction force of the earth. (Radius of the earth = 6400 km, *g* = 9.8 m/s^{2}).

An artificial satellite of mass 200 kg revolves around the earth in an orbit of average radius 6670 km. Calculate its orbital kinetic energy, the gravitational potential energy and the total energy in the orbital.

(Mass of earth = 6.0 × 10^{24} kg, *G* = 6.67 × 10^{–11} Nm^{2} kg^{ –2}).

With what velocity must a body be thrown upward from the surface of the earth so that it reaches a height of 10 *R _{e}*? Earth’s mass and

*G*= 6.67 × 10

^{ –11}Nm

^{2}kg

^{–2}.

A rocket is launched vertically from the surface of the earth with an initial velocity of 10 km s^{–1}. How far above the surface of the earth would it go? Mass of the earth = 6.0 ×^{ }10^{24} kg, radius = 6400 km and *G* = 6.67 ×^{ }10^{ –11} Nm^{2} kg^{ –2}. ^{ }

The escape velocity of a body from earth is 11.2 km s^{–1}. If the radius of a planet be half the radius of the earth and its mass be one-fourth that of earth, then what will be the escape velocity from the planet?

A particle falls on the surface of the earth from infinity. If the initial velocity of the particle is zero and friction due to air is negligible, find the velocity of the particle when it reaches the surface of the earth. Also find its kinetic energy. (Radius of earth is 6400 km and *g* is 9.8 m/s^{2}.)