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Question

Since the density of planet is same as that of earth.

$\rho_{ p }=\rho_{ e }$

$\Rightarrow \frac{ M _{ p }}{\frac{4}{3}}\pi R _{ p }^{3}=\frac{

Vedclass · Accepted Answer

$2^{\frac{1}{3}}\, {W}$

Vedclass · Answer

Since the density of planet is same as that of earth.

$\rho_{ p }=\rho_{ e }$

$\Rightarrow \frac{ M _{ p }}{\frac{4}{3}}\pi R _{ p }^{3}=\frac{ M _{ e }}{\frac{4}{3}}$

$\pi R _{ e }^{3}$

$\Rightarrow \frac{ R _{ p }}{ R _{ e }}=\left(\frac{ M _{ p }}{ M _{ e }}\right)^{1 / 3}$

The value of gravitational acceleration $=g=\frac{G M}{R^{2}}$

$\Rightarrow \frac{W_{p}}{W_{e}}=\frac{m g_{p}}{m g_{e}}$

$\Rightarrow \frac{g_{p}}{g_{e}}=\frac{M_{p}}{M_{e}} \frac{R_{e}^{2}}{R_{p}^{2}}$

$=\frac{M_{p}}{M_{e}}\left(\frac{M_{e}}{M_{p}}\right)^{2 / 3}=\left(\frac{M_{p}}{M_{e}}\right)^{1 / 3}=2^{1 / 3}$

$\Rightarrow W_{p}=2^{1 / 3}\; W$

Consider a planet in some solar system which has a mass double the mass of earth and density equal to the average density of earth. If the weight of an object on earth is ${W}$, then weight of the same object on that planet will be

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