Chapter 6, Problem 97P

To determine

The minimum speed for an object near Earth’s surface so that the object escapes both Earth’s and the Sun’s gravitational pulls.

Answer to Problem 97P

The minimum speed for an object near Earth’s surface so that the object escapes both Earth’s and the Sun’s gravitational pulls is $617.9\text{ km/s}$ .

Explanation of Solution

The total mechanical energy of the object will be constant. However, after the object escapes the both Earth’s and the Sun’s gravitational pulls, its total energy will be zero.

Write the equation for the conservation of energy of the object.

  $K_i+U_i=0$        (I)

Here, $K_i$ is the initial kinetic energy of the object and $U_i$ is the initial potential energy of the object.

Write the equation for $K_i$ .

  $K_i=\frac{1}{2}m{v}_i^2$        (II)

Here, $m$ is the mass of the object and $v_i$ is the minimum speed for an object near Earth’s surface so that the object escapes both Earth’s and the Sun’s gravitational pulls

Write the equation for $U_i$ .

  $U_i=U_{iS}+U_{iE}$        (III)

Here, $U_{iS}$ is the gravitational pull of the Sun on the object and $U_{iE}$ is the gravitational pull of the Earth on the object.

Write the equation for $U_{iS}$.

  $U_{iS}=-\frac{G{M}_{S}m}{R_S}$        (IV)

Here, $G$ is the universal gravitation constant, $M_S$ is the mass of the sun and $R_S$ is the radius of the Sun.

Write the equation for $U_{iE}$.

  $U_{iE}=-\frac{G{M}_{E}m}{R_E}$        (V)

Here, $M_E$ is the mass of the Earth and $R_E$ is the distance from the sun to the Earth.

Put equations (IV) and (V) in equation (III).

  $\begin{array}{c}{U}_i=-\frac{G{M}_{S}m}{R_S}+\left(-\frac{G{M}_{E}m}{R_E}\right)\\ =-\frac{G{M}_{S}m}{R_S}-\frac{G{M}_{E}m}{R_E}\\ =-Gm\left(\frac{M_S}{R_S}+\frac{M_E}{R_E}\right)\end{array}$        (VI)

Put equations (II) and (VI) in equation (I).

  $\begin{array}{c}\frac{1}{2}m{v}_i^2+\left[-Gm\left(\frac{M_S}{R_S}+\frac{M_E}{R_E}\right)\right]=0\\ \frac{1}{2}m{v}_i^2-Gm\left(\frac{M_S}{R_S}+\frac{M_E}{R_E}\right)=0\\ v_i^2=\frac{2Gm}{m}\left(\frac{M_S}{R_S}+\frac{M_E}{R_E}\right)\\ v_i=\sqrt{2G\left(\frac{M_S}{R_S}+\frac{M_E}{R_E}\right)}\end{array}$        (VII)

Conclusion:

The value of $G$ is $6.67\times {10}^{-11}{\text{ m}}^3\text{/kg}\cdot {\text{s}}^2$ , the mass of the sun is $1.989\times {10}^{30}\text{ kg}$ , the radius of the sun is $6.95\times {10}^5\text{ km}$ , the mass of earth is $5.972\times {10}^{24}\text{ kg}$ and the distance from the sun to the earth is $149.6\times {10}^6\text{ km}$ .

Substitute $6.67\times {10}^{-11}{\text{ m}}^3\text{/kg}\cdot {\text{s}}^2$ for $G$ , $1.989\times {10}^{30}\text{ kg}$ for $M_S$ , $6.95\times {10}^5\text{ km}$ for $R_S$ , $5.972\times {10}^{24}\text{ kg}$ for $M_E$ and $6.37\times {10}^3\text{ km}$ for $R_E$ in equation (VII) to find $v_i$.

  $\begin{array}{c}{v}_i=\sqrt{2\left(6.67\times {10}^{-11}{\text{ m}}^3\text{/kg}\cdot {\text{s}}^2\right)\left(\frac{1.989\times {10}^{30}\text{ kg}}{6.95\times {10}^5\text{ km}\frac{1000\text{ m}}{1\text{ km}}}+\frac{5.972\times {10}^{24}\text{ kg}}{149.6\times {10}^6\text{ km}\frac{1000\text{ m}}{1\text{ km}}}\right)}\\ =617.9\times {10}^3\text{ m/s}\\ =617.9\times {10}^3\text{ m/s}\frac{1\text{ km}}{1000\text{ m}}\\ =617.9\text{ km/s}\end{array}$

Therefore, the minimum speed for an object near Earth’s surface so that the object escapes both Earth’s and the Sun’s gravitational pulls is $617.9\text{ km/s}$.