91Ó°ÊÓ

radius of the planet $r=7.0×{10}^{6}m$

height of the stone traveled into crevasse $h=100m$

time taken to for stone reaching through the crevasse$t=6.0s$

The velocity of the stone$v=\frac{h}{t}=\frac{100}{6}=16.7m/s$

The initial mechanical energy of the stone, when it is located at height $h$above the the planet, is just gravitational potential energy:

$E=U=\frac{GMm}{(R+h)}$

where

$G$is the gravitational constant

$M$is the mass of the planet

$m$is the mass of the stone

$R$is the radius of the planet

$h$is the altitude of the stone

When the stone hits the ground, its mechanical energy will sum of potential energy and kinetic energy:

$E=\frac{GMm}{R}+\frac{1}{2}m{v}^2$

where

$v$is the speed of the object at the ground

Since the mechanical energy is conserved, we can write

$\frac{GMm}{R}+\frac{1}{2}m{v}^2=\frac{GMm}{R+h}$

and solving for M, we find

role="math" localid="1649932127733" $$\begin{gathered} v^2=2GM\left(\frac{1}{R+h}-\frac{1}{R}\right) \\ M=\frac{v^2}{2G\left(\frac{1}{R+h}-\frac{1}{R}\right)} \\ M=\frac{{(16.7)}^2}{2×6.67×{10}^{-11}×\left(\frac{1}{7.0×{10}^6+100}-\frac{1}{7.0×{10}^6}\right)} \\ M=\frac{139.445×{10}^{11}}{(1.42855×{10}^{-7}-1.42854×{10}^{-7})}=\frac{139.445×{10}^{11}}{1×{10}^{-12}}=1.39×{10}^{25}kg \end{gathered}$$