Question

A body of mass m is taken from the earth surface to a height equal to the radius R of earth. If ‘g’ is the acceleration due to gravity at the surface of the earth then change in the P.E of the body is​

Answer 1

g’ =

$g \frac{r}{ {(r + r)}^{2} } = g \frac{ {r}^{2} }{4 {r}^{2} } = 9.8 \times \frac{1}{4} = 2.45m {s}^{ - 2}$

There, P.E. = mgh = m × 2.45 × 2r = 4.9mr = Required answer

Explanation:

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