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At what distance (in metre) from the centre of the Moon, the intensity of gravitational field will be zero? (Take, mass of Earth and Moon as $5.98 \times 10^{24} \mathrm{~kg}$ and $7.35 \times 10^{22} \mathrm{~kg}$ respectively and the distance between Moon and Earth is $3.85 \times 10^8 \mathrm{~m}$.)
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Verified Answer
The correct answer is:
$3.85 \times 10^7$
Let X be the distance of the point from the centre of earth where gravitational intensity is zero. Therefore,
$\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{x}^2}=\frac{\mathrm{GM}_{\mathrm{m}}}{\left(3.85 \times 10^8-\mathrm{x}\right)^2}$
or $\frac{\mathrm{x}}{3.85 \times 10^8-\mathrm{x}}=\sqrt{\frac{\mathrm{M}_{\mathrm{e}}}{\mathrm{M}_{\mathrm{m}}}}=\sqrt{\frac{5.98 \times 10^{24}}{7.35 \times 10^{22}}} \cong 9$
or $\quad \frac{x}{9}+x=3.85 \times 10^8$
or $\quad \mathrm{x}=9 \times 3.85 \times 10^8 / 10=3.46 \times 10^8 \mathrm{~m}$
Distance from Moon $=3.85 \times 10^8-3.46 \times 10^8$
$=3.9 \times 10^7 \mathrm{~m} \cong 3.85 \times 10^7 \mathrm{~m}$
$\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{x}^2}=\frac{\mathrm{GM}_{\mathrm{m}}}{\left(3.85 \times 10^8-\mathrm{x}\right)^2}$
or $\frac{\mathrm{x}}{3.85 \times 10^8-\mathrm{x}}=\sqrt{\frac{\mathrm{M}_{\mathrm{e}}}{\mathrm{M}_{\mathrm{m}}}}=\sqrt{\frac{5.98 \times 10^{24}}{7.35 \times 10^{22}}} \cong 9$
or $\quad \frac{x}{9}+x=3.85 \times 10^8$
or $\quad \mathrm{x}=9 \times 3.85 \times 10^8 / 10=3.46 \times 10^8 \mathrm{~m}$
Distance from Moon $=3.85 \times 10^8-3.46 \times 10^8$
$=3.9 \times 10^7 \mathrm{~m} \cong 3.85 \times 10^7 \mathrm{~m}$
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