91Ó°ÊÓ

The escape velocityof the planet is the velocity on the surface of planet, at which if any object is thrown, then it may break out of a planet’s gravitational pull.

To calculate a planet’s escape velolity, we can use the following expression.

$V_e=\sqrt{\frac{2GM}{r}}.............\left(1\right)$

Here, G is the denotation of Universal Gravitational Constant, represent the mass of the planet and represents the planet’s radius.

From equation , the relation of escape velocity with the mass and radius of planet can be easily understood and can be written as:

$V_e\mathrm{Î\pm }\sqrt{\frac{M}{r}}...............\left(2\right)$

The escape velocity is directly proportional to the square root of the ratio of mass and radius of any planet. Higher the ratio, higher will be the escape velocity for that planet.

The mass and radius of earth are:

role="math" localid="1656483093876" $$\begin{gathered} {\mathrm{M}}_{\mathrm{E}}=6×{10}^{24}\mathrm{kg} \\ {\mathrm{r}}_{\mathrm{E}}=6.4×{10}^6\mathrm{m} \end{gathered}$$

The value of ratio will be:

$\frac{{\mathrm{M}}_{\mathrm{E}}}{{\mathrm{r}}_{\mathrm{E}}}=9.37×{10}^{17}\mathrm{kg}/\mathrm{m}$

The mass and radius of moon are:

$$\begin{gathered} {\mathrm{M}}_m=7.35×{10}^{22}\mathrm{kg} \\ {\mathrm{r}}_m=1.73×{10}^6\mathrm{m} \end{gathered}$$

The value of ratio will be:

$$\begin{gathered} \frac{{\mathrm{M}}_{\mathrm{m}}}{{\mathrm{r}}_{\mathrm{m}}}=\frac{7.35×{10}^{22}\mathrm{kg}}{1.73×{10}^6\mathrm{m}} \\ =4.25×{10}^{16}\mathrm{Kg}/\mathrm{m} \end{gathered}$$

As the value of ratio is higher for earth, the earth will have higher escape velocity.

The earth, having higher value of the ratio of mass and radius, will have higher escape velocity in comparision to moon. Thus, more fuel will be needed to escape the gravitational pull of earth.