Physics, asked by amandubey292004, 1 year ago

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

Answers

Answered by manetho
7

Answer:

7.9 x 10^2 m/s

Explanation:

Height  of the body =  Radius of Earth

Mass of Earth = 6 x 10^24 kg

Radius of Earth = 6.4 x 10^6 m

Escape velocity of earth i.e.,  V_e = (2 g R)^0.5  

First we need to find, the acceleration due to gravity at the given height.

So,    g_{h} = (9.8 x 6.4 x 10^24 )÷ (6.4 x 10^24 + 6.4 x 10^24 )            

( From, g_h = g R^2/(R+r)^2    here R=r    )

g_h = 4.9 ms^-2   ( After simplification)

Now put the value of g_h as g in (2 g R)^0.5

V_e = (2 x 4.9 x 6.4 x 10^6)^0.5

V_e ≅ 8 x 10^2  ( Exact value is 7.9 x 10^2)

velocity be it thrown so that it goes out of the gravitational field of earth = 800 m/s

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