MHT CET · Physics · Gravitation
The mass of earth is 81 times the mass of the moon and the distance between their centres is \(\mathrm{R}\). The distance from the centre of the earth where gravitational force will be zero is
- A \(\frac{9 \mathrm{R}}{10}\)
- B \(\frac{\mathrm{R}}{2}\)
- C \(\frac{\mathrm{R}}{81}\)
- D \(\frac{\mathrm{R}}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{9 \mathrm{R}}{10}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{M}_{\mathrm{E}}=81 \mathrm{M}_{\mathrm{M}}\)
Let at a distance \(x\) from the earth, the gravitational force is zero.
\(
\begin{array}{c}
\therefore \frac{G M_{E}}{x^{2}}=\frac{G M_{m}}{(R-x)^{2}} \\
\frac{G 81 M_{m}}{x^{2}}=\frac{G M_{m}}{(R-x)^{2}} \\
\frac{81}{x^{2}}=\frac{1}{(R-x)^{2}} \\
\frac{9}{x}=\frac{1}{R-x} \\
9(R-x)=x \\
9 R=10 x \\
x=\frac{9}{10} R
\end{array}
\)
Let at a distance \(x\) from the earth, the gravitational force is zero.
\(
\begin{array}{c}
\therefore \frac{G M_{E}}{x^{2}}=\frac{G M_{m}}{(R-x)^{2}} \\
\frac{G 81 M_{m}}{x^{2}}=\frac{G M_{m}}{(R-x)^{2}} \\
\frac{81}{x^{2}}=\frac{1}{(R-x)^{2}} \\
\frac{9}{x}=\frac{1}{R-x} \\
9(R-x)=x \\
9 R=10 x \\
x=\frac{9}{10} R
\end{array}
\)
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