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 ?
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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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