A body is projected from the earth's surface with the intention of letting it escape from the earth's gravitational field. What is the minimum escape velocity of the body? [Earth's radius = 6.4 x 103km, g = 10ms-2]
Question
A body is projected from the earth's surface with the intention of letting it escape from the earth's gravitational field. What is the minimum escape velocity of the body? [Earth's radius = 6.4 x 103km, g = 10ms-2]
Solution
The escape velocity of a body from the earth's gravitational field can be calculated using the formula:
v = sqrt(2gr)
where: v = escape velocity g = acceleration due to gravity r = radius of the earth
Given: g = 10 m/s^2 r = 6.4 x 10^6 m (converted from km to m)
Substituting these values into the formula, we get:
v = sqrt(2 * 10 m/s^2 * 6.4 x 10^6 m) v = sqrt(1.28 x 10^8 m^2/s^2) v = 11314 m/s
Therefore, the minimum escape velocity of the body is approximately 11314 m/s.
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