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Chapter 4: Problem 149

(a) If a spacecraft moves in a straight line between the Earth and the Sun, at what point would the force of gravity on the spacecraft due to the Sun be as large as that due to the Earth? (b) If the spacecraft is close to, but not at, this equilibrium point, does the net force on the spacecraft tend to push it toward or away from the equilibrium point? [Hint: Imagine the spacecraft a small distance \(d\) closer to the Earth and find out which gravitational force is stronger.]

Short Answer

Expert verified
Answer: The equilibrium distance is given by the formula \(r_e = \frac{m_ER^2}{2m_ER + (m_E + m_S)r_e}\), where \(r_e\) is the distance from the Earth to the spacecraft, \(m_E\) is Earth's mass, \(m_S\) is the Sun's mass, and \(R\) is the distance between Earth and the Sun. When the spacecraft is slightly closer to the Earth, the net force pushes it away from the Earth and towards the equilibrium point.

Step by step solution

01

Write the equations for gravitational force

Let \(F_e\) be the gravitational force due to the Earth on the spacecraft and \(F_s\) be the gravitational force due to the Sun on the spacecraft. Using the gravitational force formula, we have: \(F_e = G\frac{m_Em}{r_e^2}\) (Earth's gravitational force) \(F_s = G\frac{m_Sm}{r_s^2}\) (Sun's gravitational force) Where \(m_E\) is Earth's mass, \(m_S\) is the Sun's mass, \(m\) is the spacecraft's mass, \(r_e\) is the distance from the Earth to the spacecraft, and \(r_s\) is the distance from the Sun to the spacecraft.
02

Set up the equilibrium condition

The equilibrium condition is that at a certain point, the gravitational force due to the Earth equals the gravitational force due to the Sun. Therefore, we have: \(F_e = F_s\) Plugging in the equations from Step 1, we get: \(G\frac{m_Em}{r_e^2} = G\frac{m_Sm}{r_s^2}\)
03

Solve for the distance from the Earth

To find the equilibrium point, we need to find the distance \(r_e\) from the Earth when the forces are equal. We can simplify the equation from Step 2 by canceling out like terms: \(\frac{m_E}{r_e^2} = \frac{m_S}{r_s^2}\) We know that \(r_e + r_s\) is the distance between Earth and the Sun, which we can denote as \(R\). So, \(r_s = R - r_e\). Now, we substitute \(r_s\) in the equation: \(\frac{m_E}{r_e^2} = \frac{m_S}{(R - r_e)^2}\) Now, we can solve for \(r_e\): \(m_ER^2 - 2m_ERr_e + m_Er_e^2 = m_Sr_e^2\)
04

Find the equilibrium distance

Solve for \(r_e\) to find the equilibrium distance: \(r_e = \frac{m_ER^2}{2m_ER + (m_E + m_S)r_e}\)
05

Analyze the net force when the spacecraft is close to the equilibrium point

Let the spacecraft be a small distance \(d\) closer to the Earth. The gravitational force due to the Earth becomes stronger, while the gravitational force due to the Sun becomes weaker. This means that \(F_e > F_s\). Since the Earth's gravitational force is stronger, the net force on the spacecraft pushes it away from the Earth and towards the equilibrium point.

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