Solve the integral I - -RGMmx2dx-

I = ∫_∞^RGMmx^2dx = GMm ∫_∞^Rdxx^2 = GMm [ 1x ]_∞^R = GMm [ 1R + 1∞ ] = GMmR The above expression represents gravitational potential energy which i ...

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Standard XII

Physics

Definite Integrals

Solve the int...

Question

Solve the integral $I = \int_{\infty}^{R} \frac{G M m}{x^{2}} d x$ .

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Solution

$I = \int_{\infty}^{R} \frac{G M m}{x^{2}} d x = G M m \int_{\infty}^{R} \frac{d x}{x^{2}} = G M m {[- \frac{1}{x}]}_{\infty}^{R}$
$= G M m [- \frac{1}{R} + \frac{1}{\infty}] = \frac{- G M m}{R}$
The above expression represents gravitational potential energy which is obtained by integrating $G M m / x^{2}$ between the limits infinity $(\infty)$ and radius of the earth $(R)$ .

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Solve the integral I=∫R∞GMmx2dx.

I=∫R∞GMmx2dx=GMm∫R∞dxx2=GMm[−1x]R∞=GMm[−1R+1∞]=−GMmRThe above expression represents gravitational potential energy which is obtained by integrating GMm/x2 between the limits infinity (∞) and radius of the earth (R).

Solve the integral $I = \int_{\infty}^{R} \frac{G M m}{x^{2}} d x$ .