91Ó°ÊÓ

The given data can be listed below as,

• The radius of planet is$r=3.24×{10}^{6}m$.
• The escape velocity of planet is,localid="1657178488729" $\mathrm{v}=7.65×{10}^3\mathrm{m}/\mathrm{s}$

The term acceleration due to gravity is the rate of speed at which an object will exceed the earth's rotation per second that will cause it to fall down is the effect of gravitational pull towards the earth.

From conservation law of energy,

$K_1+U_1=K_2+U_2$

Here,$K_1$is the kinetic energy at one point,$U_1$is the potential energy at one point,$K_2$is the kinetic energy at another point,$U_2$is the potential energy at another point.

At the escape speed, the object has no kinetic energy when it is very far away from the planet. And r is infinite so potential energy at that point is also become zero.

So above equation will become,

$K_1+U_1=0+0$

Substitute the value of kinetic and potential energy.

$\frac{1}{2}m{v}^2-\frac{GMm}{r}=0$

Here, Mis the mass of the planet,G is theconstant of gravitation whose value is$\mathrm{G}=6.673×{10}^{-11}\mathrm{N}.{\mathrm{m}}^2/{\mathrm{kg}}^2$,m is the mass of the object, r is the radius of planet and v is the escape velocity of the planet.

Solving the above equation for M,

$M=\frac{r{v}^2}{2G}$

Acceleration due to gravity can be expressed as,

$g=\frac{GM}{r^2}$

Substitute all the values in the above,

localid="1657178592381" $$\begin{gathered} g=\frac{G\left({\displaystyle \frac{r{v}^2}{2G}}\right)}{r^2} \\ =\frac{v^2}{2r} \\ =\frac{{\left(7.65×{10}^3\mathrm{m}/\mathrm{s}\right)}^2}{2\left(3.24×{10}^6\mathrm{m}\right)} \\ =9.03\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the acceleration due to gravity is localid="1657178602816" $9.03\mathrm{m}/{\mathrm{s}}^2$.