A body weighs $48\text{ N}$ on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is:
A
16 N
B
27 N
C
32 N
D
36 N
The answer is B.
Solution
At height $h = R/3$, distance from centre is $r = 4R/3$. Gravity varies as $1/r^2$, so $F = 48 \times (3/4)^2 = 48 \times 9/16 = 27\text{ N}$.
Gravitational force varies inversely with the square of the distance from the Earth's centre: $F(r) = \frac{G M m}{r^2}$. At the surface, $r_1 = R$ and $F_1 = 48\text{ N}$. At height $h = R/3$, the radial distance is $r_2 = R + \frac{R}{3} = \frac{4R}{3}$. The new gravitational force is $F_2 = F_1 \left(\frac{R}{r_2}\right)^2 = 48 \times \left(\frac{R}{4R/3}\right)^2 = 48 \times \left(\frac{3}{4}\right)^2 = 48 \times \frac{9}{16} = 3 \times 9 = 27\text{ N}$.
ConceptVariation of Gravity with Height
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66 questions from this chapter have appeared across 26 exam years.
