Chapter 4, Problem 155P

(a)

To determine

The point at which the force of gravity on the spacecraft due to the Sun is as large as that due to the Earth.

Answer to Problem 155P

The point at which the force of gravity on the spacecraft due to the Sun is as large as that due to the Earth is $\underset{\_}{2.60\times {10}^8\text{m from Earth}}$.

Explanation of Solution

Write the expression for the magnitude of gravitational force of Sun on the spacecraft.

$F_{\text{sS}}=\frac{G{M}_{\text{S}}m}{r_{\text{S}}^2}$ (I)

Here, $F_{\text{sS}}$ is the gravitational force of Sun on the spacecraft, $M_{\text{S}}$ is the mass of Sun, $m$ is the mass of spacecraft, $r_{\text{S}}$ is the distance between spacecraft and Sun, and $G$ is the gravitational constant.

Write the expression for the magnitude of gravitational force of Earth on the spacecraft.

$F_{\text{sE}}=\frac{G{M}_{\text{E}}m}{r_{\text{E}}^2}$ (II)

Here, $F_{\text{sE}}$ is the gravitational force of Earth on the spacecraft, $M_{\text{E}}$ is the mass of Earth, and $r_{\text{E}}$ is the distance between spacecraft and Earth.

Equate the right hand sides of equations (I) and (II) since the forces are to be equal, and solve for $r_{\text{E}}$.

$\begin{array}{c}\frac{G{M}_{\text{S}}m}{r_{\text{S}}^2}=\frac{G{M}_{\text{E}}m}{r_{\text{E}}^2}\\ \frac{r_{\text{E}}^2}{r_{\text{S}}^2}=\frac{M_{\text{E}}}{M_{\text{S}}}\\ r_{\text{E}}=r_{\text{S}}\sqrt{\frac{M_{\text{E}}}{M_{\text{S}}}}\end{array}$ (III)

Write the expression for $r_{\text{S}}$ in terms of the Earth-Sun distance ($D$).

$r_{\text{S}}=D-r_{\text{E}}$ (IV)

Use equation (IV) in (III).

$\begin{array}{c}{r}_{\text{E}}=\left(D-r_{\text{E}}\right)\sqrt{\frac{M_{\text{E}}}{M_{\text{S}}}}\\ r_{\text{E}}\sqrt{\frac{M_{\text{S}}}{M_{\text{E}}}}=\left(D-r_{\text{E}}\right)\\ r_{\text{E}}=\frac{D}{\left(1+\sqrt{\frac{M_{\text{S}}}{M_{\text{E}}}}\right)}\end{array}$ (V)

The mass of Sun is $1.987\times {10}^{30}\text{kg}$, the mass of Earth is $5.974\times {10}^{24}\text{kg}$, and the Earth-Sun distance is $1.50\times {10}^{11}\text{m}$.

Conclusion:

Substitute $1.50\times {10}^{11}\text{m}$ for $D$, $1.987\times {10}^{30}\text{kg}$ for $M_{\text{S}}$, and $5.974\times {10}^{24}\text{kg}$ for $M_{\text{E}}$ in equation (V) to find $r_{\text{E}}$.

$\begin{array}{c}{r}_{\text{E}}=\frac{1.50\times {10}^{11}\text{m}}{\left(1+\sqrt{\frac{1.987\times {10}^{30}\text{kg}}{5.974\times {10}^{24}\text{kg}}}\right)}\\ =2.60\times {10}^8\text{m}\end{array}$

Therefore, the point at which the force of gravity on the spacecraft due to the Sun is as large as that due to the Earth is $\underset{\_}{2.60\times {10}^8\text{m from Earth}}$.

(b)

To determine

Whether the net force on the spacecraft tend to push it toward or away from equilibrium point when the spacecraft is close to, but not at the equilibrium point.

Answer to Problem 155P

The net force on the spacecraft tend to push it $\underset{\_}{\text{away from}}$ equilibrium point when the spacecraft is close to, but not at the equilibrium point.

Explanation of Solution

At the equilibrium point, the gravitational force on the spacecraft due to the Sun and the Earth are equal in magnitude but opposite in direction. Thus, the net gravitational force is zero at that point. When the spacecraft is close to the Earth than at equilibrium point, the force due to Earth is greater than that due to the Sun. Similarly, when the spacecraft is close to the Sun than at equilibrium point, the force due to Sun is greater than that due to the Earth.

In both cases, if the spacecraft is close to, but not at the equilibrium point, then it will be pulled away from the equilibrium point such that it falls towards the stronger field source.

Conclusion:

Therefore, the net force on the spacecraft tend to push it $\underset{\_}{\text{away from}}$ equilibrium point when the spacecraft is close to, but not at the equilibrium point.