The mass of the moon is (1/8) of the earth bu | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$\frac{\mathrm{g}_{\mathrm{m}}}{\mathrm{g}_{\mathrm{e}}}=\frac{\frac{\mathrm{G}(\mathrm{Me} / 8)}{\mathrm{R}_{\mathrm{M}}^{2}}}{\frac{\mathrm{GMe}}{\m

Vedclass · Accepted Answer

the radius of the earth is $\left( {\sqrt {8/6} } \right)$ of the moon

Vedclass · Answer

$\frac{\mathrm{g}_{\mathrm{m}}}{\mathrm{g}_{\mathrm{e}}}=\frac{\frac{\mathrm{G}(\mathrm{Me} / 8)}{\mathrm{R}_{\mathrm{M}}^{2}}}{\frac{\mathrm{GMe}}{\mathrm{R}_{\mathrm{e}}^{2}}}=\frac{\mathrm{R}_{\mathrm{e}}^{2}}{\mathrm{8R}_{\mathrm{M}}^{2}}=\frac{1}{6} \Rightarrow \mathrm{R}_{\mathrm{e}}=\sqrt{\frac{8}{6} \mathrm{R}_{\mathrm{M}}}$

The mass of the moon is $(1/8)$ of the earth but the gravitational pull is $(1/6)$ of the earth. It is due to the fact that.

Similar Questions

Derive the equation for variation of $g$ due to height from the surface of earth.

If $R$ is the radius of the earth and the acceleration due to gravity on the surface of earth is $g=\pi^2 \mathrm{~m} / \mathrm{s}^2$, then the length of the second's pendulum at a height $h=2 R$ from the surface of earth will be,:

If the earth stops rotating, the value of $‘g’$ at the equator will

Assuming earth to be a sphere of a uniform density, what is the value of gravitational acceleration in a mine $100\, km$ below the earth’s surface ........ $m/{s^2}$. (Given $R = 6400 \,km$)

The mass of moon is $7.34 \times {10^{22}}\,kg$ and radius of moon is $1.74 \times {10^6}\,m$ then The value of gravitation accelaration will be ....... $N/kg$