Question:

Knowing that mass of Moon is $\frac{M}{81}$ where $M$ is the mass of Earth, find the distance of the point where gravitational field due to Earth and Moon cancel each other, from the Moon. Given that distance between Earth and Moon is $60\, R$, where $R$ is the radius of Earth.

Updated On: Jul 5, 2022
  • 2 R
  • 4 R
  • 6 R
  • 8 R
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The Correct Option is C

Solution and Explanation

Let C be the point where the gravitational field due to earth and the moon be equal. C is at a distance x from the moon and (60R - x) from the earth. We know that the gravitational field at a distance r from mass M is given by $E=G\frac{M}{r^{2}}$ In our case E = E $G \frac{M}{\left(60R-x\right)^{2}}=\frac{M}{G\frac{81}{x^{2}}}$ $\Rightarrow 81x^{2} = \left(60R - x\right)^{2}$ $\Rightarrow \left(9x\right)^{2} = \left(60R - x\right)^{2}$ $\therefore 9x = \pm \left(60R - x\right) \therefore 9x = 60R - x or 9x = x - 60R$ $\Rightarrow 10x = 60R or 8x = -60R$ $\Rightarrow x = 6R$ or $x=-\frac{60}{8}R$ which is not possible as distance cannot be negative.
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Concepts Used:

Gravitation

In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.

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,

  • F ∝ (M1M2) . . . . (1)
  • (F ∝ 1/r2) . . . . (2)

On combining equations (1) and (2) we get,

F ∝ M1M2/r2

F = G × [M1M2]/r2 . . . . (7)

Or, f(r) = GM1M2/r2

The dimension formula of G is [M-1L3T-2].