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Chapter 4: Problem 133

The mass of the Moon is 0.0123 times that of the Earth. A spaceship is traveling along a line connecting the centers of the Earth and the Moon. At what distance from the Earth does the spaceship find the gravitational pull of the Earth equal in magnitude to that of the Moon? Express your answer as a percentage of the distance between the centers of the two bodies.

Short Answer

Expert verified
Answer: Approximately 93.80%

Step by step solution

01

Given variables

We are given the mass ratio between the Moon and Earth, which is \(0.0123\). Let's call the mass of the Earth \(M_E\) and the mass of the Moon \(M_M\), then \(M_M = 0.0123M_E\). We are also asked to find the distance where gravitational pull is equal, let's call it \(x\), from Earth.
02

Formula for gravitational force

The formula for calculating the gravitational force \(F\) acting on a spaceship between two celestial bodies with masses \(m_1\) and \(m_2\) at a distance \(d\) from each other is given by: $$F = G \cdot \frac{m_1 \cdot m_2}{d^2}$$ where \(G\) is the gravitational constant given by \(6.674 \times 10^{-11} \, N \cdot (m/kg)^2\).
03

Set up the equation for Earth's and Moon's gravitational pull

Let's denote the spaceship mass as \(m\). At the point we are trying to find, the gravitational forces acting on the spaceship due to Earth and the Moon should be equal: $$F_E = F_M$$ Using the gravitational force formula: $$G \cdot \frac{M_E \cdot m}{x^2} = G \cdot \frac{M_M \cdot m}{(D-x)^2}$$ where \(F_E\) is the force exerted by Earth, \(F_M\) is the force exerted by the Moon, \(D\) is the total distance between the centers of the Earth and the Moon. Since we are only interested in the distance, we can divide both sides by \(Gm\) and substitute \(M_M\) with \(0.0123M_E\): $$\frac{M_E}{x^2} = \frac{0.0123M_E}{(D-x)^2}$$
04

Solve for x

We will first eliminate the \(M_E\) from both sides by dividing both sides by \(M_E\): $$\frac{1}{x^2} = \frac{0.0123}{(D-x)^2}$$ Now, let's cross-multiply: $$(D-x)^2 = 0.0123x^2$$ Expand the left side by multiplying \((D-x)(D-x)\): $$D^2 - 2Dx + x^2 = 0.0123x^2$$ Now, let's move all terms to the left side: $$D^2 -2Dx + (1 - 0.0123)x^2 = 0$$
05

Express x as a percentage of the total distance

Let's denote \(p\) the percentage of distance x as a percentage of total distance D. Then: $$x = \frac{p}{100}D$$ We'll now substitute this into the previous equation: $$D^2 - 2D\left(\frac{p}{100}D\right) + (1 - 0.0123)\left(\frac{p}{100}D\right)^2 = 0$$ Divide both sides of the equation by \(D^2\): $$1 - 2\frac{p}{100} + (1 - 0.0123)\left(\frac{p}{100}\right)^2 = 0$$ Now, we can solve this quadratic equation for percentage \(p\). The numerical result is approximately: $$p \approx ≈ 93.80 \%$$ Therefore, the location where the spaceship experiences equal gravitational pull from Earth and the Moon is approximately \(93.80 \%\) of the distance between the centers of the two bodies.

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Most popular questions from this chapter

A block of mass \(m_{1}=3.0 \mathrm{kg}\) rests on a frictionless horizontal surface. A second block of mass \(m_{2}=2.0 \mathrm{kg}\) hangs from an ideal cord of negligible mass that runs over an ideal pulley and then is connected to the first block. The blocks are released from rest. (a) Find the acceleration of the two blocks after they are released. (b) What is the velocity of the first block 1.2 s after the release of the blocks, assuming the first block does not run out of room on the table and the second block does not land on the floor? (c) How far has block 1 moved during the 1.2 -s interval? (d) What is the displacement of the blocks from their initial positions 0.40 s after they are released?
The coefficient of static friction between block \(\mathrm{A}\) and a horizontal floor is 0.45 and the coefficient of static friction between block \(\mathrm{B}\) and the floor is \(0.30 .\) The mass of each block is $2.0 \mathrm{kg}$ and they are connected together by a cord. (a) If a horizontal force \(\overrightarrow{\mathbf{F}}\) pulling on block \(\mathrm{B}\) is slowly increased, in a direction parallel to the connecting cord, until it is barely enough to make the two blocks start moving, what is the magnitude of \(\overrightarrow{\mathbf{F}}\) at the instant that they start to slide? (b) What is the tension in the cord connecting blocks \(A\) and \(B\) at that same instant?
A box full of books rests on a wooden floor. The normal force the floor exerts on the box is \(250 \mathrm{N}\). (a) You push horizontally on the box with a force of \(120 \mathrm{N},\) but it refuses to budge. What can you say about the coefficient of static friction between the box and the floor? (b) If you must push horizontally on the box with a force of at least \(150 \mathrm{N}\) to start it sliding, what is the coefficient of static friction? (c) Once the box is sliding, you only have to push with a force of \(120 \mathrm{N}\) to keep it sliding. What is the coefficient of kinetic friction?
By what percentage does the weight of an object change when it is moved from the equator at sea level, where the effective value of \(g\) is $9.784 \mathrm{N} / \mathrm{kg},\( to the North Pole where \)g=9.832$ N/kg?
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