Question:
Three uniform spheres of mass M and radius R earth are kept in such a way that each touches the other two. The magnitude of the gravitational force on any of the spheres due to the other two is
Three uniform spheres of mass M and radius R earth are kept in such a way that each touches the other two. The magnitude of the gravitational force on any of the spheres due to the other two is
Updated On: Jul 7, 2022
- $\frac{\sqrt3}{4}\frac{GM^2}{R^2}$
- $\frac{3}{2}\frac{GM^2}{R^2}$
- $\frac{\sqrt3GM^2}{R^2}$
- $\frac{\sqrt3}{2}\frac{GM^2}{R^2}$
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The Correct Option is A
Solution and Explanation
Force between any two spheres will be
$F =\frac{G(M) (M) }{(2R)^2} = \frac{GM^2}{4R^2}$
The two forces of equal magnitude F are acting at angle $60^\circ$ on any o f the sphere
$F_{net} = \sqrt{F^2 + F^2 + 2(F) (F) \, cos \, 60^\circ} = \sqrt{3}F$
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Concepts Used:
Newtons Law of Gravitation
Gravitational Force
Gravitational force is a central force that depends only on the position of the test mass from the source mass and always acts along the line joining the centers of the two masses.
Newton’s Law of Gravitation:
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- Directly proportional to the product of their masses i.e. F ∝ (M1M2) . . . . (1)
- Inversely proportional to the square of the distance between their center i.e. (F ∝ 1/r2) . . . . (2)
By combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2 [f(r)is a variable, Non-contact, and conservative force]