Chapter 4, Problem 50Q

To determine

The escape speed from the surfaces of Earth and Mars. State using these values whether it is easier for a rock to escape from Mars or Earth. Use the values provided in Appendix 2.

Answer to Problem 50Q

Solution:

The escape speed from the surface of Earth is 11.2 km/s and that from Mars is 5 km/s. It would be easier for the rock to escape from Mars.

Explanation of Solution

Given data:

Refer to Appendix 2 for the following values:

The equatorial diameter of Earth is $12756\text{km}$ and that of Mars is $6794\text{km}$.

The mass of Earth is $5.974\times {10}^{24}\text{ kg}$ and that of Mars is $6.418\times {10}^{23}\text{ kg}$.

Formula used:

State the expression for escape velocity.

$V=\sqrt{\frac{2GM}{R}}$

Here, $V$ is the escape velocity, $G$ is the constant of gravitation, $M$ is the mass of the planet, and $R$ is the radius of the planet.

Explanation:

It is known that the radius of a sphere is half of its diameter. Consider Earth to be spherical in shape and calculate its radius as,

$R_e=\frac{D_e}{2}$

Here, subscript e refers to the corresponding quantities for Earth.

Substitute $12756\text{km}$ for $D_e$.

$\begin{array}{c}{R}_e=\frac{12756\text{km}}{2}\\ =6378\text{ km}\left(\frac{1000\text{ m}}{1\text{ km}}\right)\\ =6378000\text{ m}\end{array}$

Consider Mars to be spherical in shape and calculate its radius as,

$R_m=\frac{D_m}{2}$

Here, subscript m refers to the corresponding quantities for Mars.

Substitute $6794\text{km}$ for $D_m$.

$\begin{array}{c}{R}_m=\frac{6794\text{km}}{2}\\ =3397\text{ km}\left(\frac{1000\text{ m}}{1\text{ km}}\right)\\ =3397000\text{ m}\end{array}$

State the expression for the escape velocity for Earth.

$V_e=\sqrt{\frac{2G{M}_e}{R_e}}$

Here, subscript e refers to the corresponding quantities for Earth.

Substitute $6.67\times {10}^{-11}{\text{m}}^3\cdot {\text{kg}}^{-1}\cdot {\text{s}}^{-2}$ for $G$, $6378000\text{ m}$ for $R_e$, and $5.974\times {10}^{24}\text{ kg}$ for $M_e$.

$\begin{array}{c}{V}_e=\sqrt{\frac{\left(2\right)\left(6.67\times {10}^{-11}{\text{m}}^3\cdot {\text{kg}}^{-1}\cdot {\text{s}}^{-2}\right)\left(5.974\times {10}^{24}\text{ kg}\right)}{\left(6378000\text{ m}\right)}}\\ =11178\text{m/s}\left(\frac{1\text{ km/s}}{1000\text{ km/s}}\right)\\ \simeq 11.2\text{ km/s}\end{array}$

State the expression for the escape velocity for Mars.

$V_m=\sqrt{\frac{2G{M}_m}{R_m}}$

Here, subscript m refers to the corresponding quantities for Mars.

Substitute $6.67\times {10}^{-11}{\text{m}}^3\cdot {\text{kg}}^{-1}\cdot {\text{s}}^{-2}$ for $G$, $3397000\text{ m}$ for $R_m$, and $6.418\times {10}^{23}\text{ kg}$ for $M_m$.

$\begin{array}{c}{V}_m=\sqrt{\frac{\left(2\right)\left(6.67\times {10}^{-11}{\text{m}}^3\cdot {\text{kg}}^{-1}\cdot {\text{s}}^{-2}\right)\left(6.418\times {10}^{23}\text{ kg}\right)}{\left(3397000\text{ m}\right)}}\\ =5020.3\text{m/s}\left(\frac{1\text{ km/s}}{1000\text{ km/s}}\right)\\ \simeq 5\text{ km/s}\end{array}$

Conclusion:

Since the escape speed from the surface of Mars is lower than that from Earth, it would be easier for the rock to escape from Mars.