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Chapter 9: Problem 18

If the distance between centres of earth and moon is \(D\) and the mass of earth is 81 times the mass of moon, then at what distance from centre of earth the gravitational force will be zero (a) \(D / 2\) (b) \(2 D / 3\) (c) \(4 D / 3\) (d) \(9 D / 10\)

Short Answer

Expert verified
The gravitational force will be zero at a distance of approximately \((9 / 10)D\) from the centre of the Earth, making the correct answer (d).

Step by step solution

01

Define variables

The gravitational force from both the Earth and the Moon must be equal and opposite at the location we're trying to find. If \(d\) is the distance from the Earth where the gravitational force is zero, and \(M_e\) and \(M_m\) represent the mass of the Earth and the Moon respectively, then the gravitational forces from Earth and the Moon are \(F_e = G(M_e / d^2)\) and \(F_m = G(M_m / (D - d)^2)\) respectively, where \(G\) is the gravitational constant. We know that \(M_e = 81M_m\) from the question.
02

Equate Forces

We can equate the forces \(F_e\) and \(F_m\) because they must be equal and opposite for the overall force to be zero. This gives us the equation \( G(M_e / d^2) = G(M_m / (D - d)^2)\). We can cancel out \(G\), as it appears on both sides of the equation, so our equation is now simplified to \(M_e / d^2 = M_m / (D - d)^2\).
03

Substitute and Solve

Now, substitute \(M_e = 81M_m\) into our equation to give \((81M_m) / d^2 = M_m / (D - d)^2\). We can simplify this equation to \(81 / d^2 = 1 / (D - d)^2\). The cross multiplication of this equation gives \(81(D - d)^2 = d^2\), which we simplify to \(81D^2 - 162Dd + 81d^2 = d^2\). Solving this equation for \(d\) gives \(d = 81D / 82\), which is approximately \((9 / 10)D\).

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