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The change in potential energy when a body of mass m is raised to a height 3r from the surface of earth will be (r-radius of earth)

a. mgR n/(n-1)
b. mgR
c. mgR n/(n+1)
d. mgR n²/(n²+1)

User JonasV
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1 Answer

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Final answer:

The change in potential energy as a body is raised to a height three times the radius of the Earth is represented by the formula ΔU = (3/4)mgr, which is derived using the law of universal gravitation and the value of acceleration due to gravity on Earth. Option number b is correct.

Step-by-step explanation:

The change in potential energy when a body of mass m is raised to a height 3r (where r is the radius of Earth) can be calculated using the law of gravitation.

The formula for the gravitational potential energy U at a distance r from the center of the Earth is U = -GMm/r, where G is the gravitational constant, M is the mass of the Earth, and m is the mass of the object. When the object is raised from the Earth's surface to a height of 3r, we can express the change in potential energy (ΔU) as the difference in potential energy at 4r and r, which gives us ΔU = GMm/r - GMm/4r = GMm(1/r - 1/4r) = (3/4)GMm/r.

Substituting the value of g (acceleration due to gravity) as GM/r2, the equation simplifies to ΔU = (3/4)mgr, which corresponds to the potential energy change.

User Guy Goldstein
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