Question

4) r" 03. The escape speed of a body from the surface of earth (radius of earth = R,) is​

Answer 1

Explanation:

The escape velocity from the surface of the earth is

v e= 2GM/RE . . . . .(1)

Acceleration due to gravity,

g= R EE2/GM

R E

GG=gR E . . . . . . . . . .(2)

Substitute equation (2) in equation (1), we get

v e =√ 2gRE,,

Answer 2

Explanation:

The acceleration due to gravity (earth), g = 9.8 m/s².

The radius (earth), R = 6.4 × 106 m.

The escape velocity (earth), Ve = √2 × 9.8 × 6.4 × 106.

Therefore, Ve = 11.2 × 103 m/s = 11.186 km/s or 11.2 km/s (Approximately).

The escape speed of the earth at the surface is approximately 11.2 km/s. This means to escape from earth’s gravity and travel to infinite space, an object must have a minimum of 11.2 km/s of the initial velocity.