Question

A particle is launched from a height 8R above the surface of earth and it is given a speed \[\left( {\dfrac{{GM}}{{8R}}} \right)\] parallel to the surfaces then
A. It will escape from the gravitational field of earth.
B. It will follow a circular orbit.
C. It will follow an elliptical orbit.
D. the point of launching will be the perigee of its orbit.

Answer 1

Hint: On the surface of the earth escape velocity is defined as the minimum velocity which is given to a body to make it free from the earth.

Formula used:
The formula for escape velocity is given as,
\[v = \sqrt {\dfrac{{2GM}}{R}} \]
Where v is the escape velocity
G is the gravitational constant
M is the mass of the planet
R is the radius

Complete step by step solution:
Given the speed of a particle is \[\left( {\dfrac{{GM}}{{8R}}} \right)\]. As we know the escape velocity is defined as the minimum velocity required by an object to escape the gravitational field without ever falling back. An object that has the escape velocity at the earth’s surface will escape the earth’s gravitational field if ignoring the losses due to the atmosphere. The escape velocity is given as,
\[v = \sqrt {\dfrac{{2GM}}{R}} \\ \]
Now if we compare both the velocity, we find that the speed is more than the orbital but less than the escape velocity therefore an orbit will form. The speed is perpendicular to the radius. It must be either an apogee or a perigee. But since the speed will be increasing, the launch point must be the perigee.

Hence option D is the correct answer.

Note: Escape velocity from the surface of the earth is given as 11.2 km/s (which is approximately equal to 6.96 miles. The value of escape velocity depends on the radius of the planet and the mass of the planet from which it is being projected. And the value of escape velocity does not depend upon the mass of the projected body and also does not depend on the angle and direction of projection. The escape velocity from the less massive moon is about 2.4 km per second at the surface of the earth.