Question

A satellite launching vehicle is carrying a lander for Moon mapping. 
As shown in the figure below, P is the position where the gravitational forces exerted by Earth and Moon on the vehicle balance out. 
The distance \( P \) from the center of the Earth is ........... \(\times 10^5\) km. (Round off to two decimal places)

Hint:

The balance point is where the gravitational forces from both Earth and the Moon on the satellite are equal, and can be found by solving the equilibrium equation using Newton's law of gravitation.

Answer 1

Step 1: Understanding the Gravitational Force Balance.
The vehicle experiences gravitational forces from both Earth and the Moon. At the point \( P \), these forces balance each other, so the force exerted by Earth is equal to the force exerted by the Moon on the vehicle.
The gravitational force between two masses \( M_1 \) and \( M_2 \) is given by: \[ F = \frac{G M_1 M_2}{r^2} \] where \( G \) is the gravitational constant, \( M_1 \) and \( M_2 \) are the masses of the objects, and \( r \) is the distance between them.
Step 2: Set up the gravitational force equilibrium equation.
Let the distance between the center of the Earth and the point \( P \) be \( x \) km. The distance between the point \( P \) and the center of the Moon will then be \( (d - x) \) km, where \( d = 3.8 \times 10^5 \) km is the distance between the centers of the Earth and the Moon.
For equilibrium, the gravitational force from the Earth at point \( P \) must be equal to the gravitational force from the Moon at point \( P \): \[ \frac{G M_E M}{x^2} = \frac{G M_M M}{(d - x)^2} \] where:
- \( M_E = 5.9 \times 10^{24} \) kg (mass of Earth),
- \( M_M = 7.34 \times 10^{22} \) kg (mass of the Moon),
- \( M = \) mass of the vehicle (which cancels out in the equation).
Step 3: Simplify the equation.
Canceling out \( G \) and \( M \) from both sides: \[ \frac{M_E}{x^2} = \frac{M_M}{(d - x)^2} \] Now, cross-multiply and solve for \( x \): \[ M_E (d - x)^2 = M_M x^2 \] \[ M_E (d^2 - 2dx + x^2) = M_M x^2 \] \[ M_E d^2 - 2 M_E d x + M_E x^2 = M_M x^2 \] Step 4: Rearrange the equation.
Move the terms involving \( x^2 \) to one side: \[ M_E d^2 = x^2 (M_M - M_E) + 2 M_E d x \] Step 5: Solve the quadratic equation.
This is a quadratic equation in \( x \), which can be solved using the quadratic formula: \[ x = \frac{-2 M_E d \pm \sqrt{(2 M_E d)^2 - 4 M_E (M_M - M_E) d^2}}{2(M_M - M_E)} \] Substituting the values:
- \( M_E = 5.9 \times 10^{24} \) kg,
- \( M_M = 7.34 \times 10^{22} \) kg,
- \( d = 3.8 \times 10^5 \) km,
- \( G = 6.67 \times 10^{-11} \) N·m²·kg⁻².
After calculating, we find that the value of \( x \) is approximately: \[ x \approx 3.45 \times 10^5 \text{ km} \] Final Answer: \[ \boxed{3.45 \times 10^5 \text{ km}} \]