First slide

Variation of g

Question

A bodyweighs 72 Non the surface of the earth. What is the gravitational force on it due to earth at a height equal to half the radius of the earth from the surface ?

Easy

A

16 N
B

28 N
C

32 N
D

72 N

Solution

Let W₁ be the weight of the body on the surface of the earth. If W₂ be the gravitational force on the body due to earth at height R₂ above the surface of the earth, then
$\frac{W_{1}}{W_{2}} = \frac{g_{1}}{g_{2}} = \frac{R_{2}^{2}}{R_{1}^{2}}$
where R₁ =radius of the earth.
Here R₁ = R and R₂ $= (R + \frac{R}{2})$
$∴ \frac{W_{1}}{W_{2}} = \frac{(R + \frac{R}{2})}{R^{2}} = \frac{9}{4}$
Or $W_{2} = W_{1} \times (\frac{4}{9}) = 72 \times (\frac{4}{9}) = 32 N$

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