A uniform thin rod of mass m and length R is placed normally on surface of earth as shown. The mass of earth is M and its radius is R . Then the magnitude of gravitational force exerted by earth on th

Question

A uniform thin rod of mass m and length R is placed normally on surface of earth as shown. The mass of earth is M and its radius is R . Then the magnitude of gravitational force exerted by earth on the rod is X R 2 3 GM m ​ then find the value of X ?

TestHub · Accepted Answer

To find the gravitational force, we consider a differential element $d x$ of the rod at a distance $x$ from the center of the Earth. The mass of this element is $d m = \frac{m}{R} d x$ .

The gravitational force $d F$ on this element is given by Newton's Law of Gravitation:

$d F = \frac{GM d m}{x ^{2}} = \frac{GM ( \frac{m}{R} d x )}{x ^{2}}$

The rod extends from the Earth's surface ( $x = R$ ) to a distance $R + R = 2 R$ from the center of the Earth.

So, we integrate $d F$ from $x = R$ to $x = 2 R$ :

$F = \int_{R}^{2 R} \frac{GM m}{R x ^{2}} d x$

$F = \frac{GM m}{R} \int_{R}^{2 R} x^{- 2} d x$

$F = \frac{GM m}{R} [- \frac{1}{x}]_{R}^{2 R}$

$F = \frac{GM m}{R} (- \frac{1}{2 R} - (- \frac{1}{R}))$

$F = \frac{GM m}{R} (\frac{1}{R} - \frac{1}{2 R})$

$F = \frac{GM m}{R} (\frac{2 - 1}{2 R})$

$F = \frac{GM m}{R} (\frac{1}{2 R})$

$F = \frac{GM m}{2 R ^{2}}$

The problem states the force is $\frac{3 GM m}{X R ^{2}}$ .

Comparing our result with the given form:

$\frac{GM m}{2 R ^{2}} = \frac{3 GM m}{X R ^{2}}$

$1/2 = 3/ X$

$X = 6$

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