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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: $$p \approx 鈮� 93.80 \%$$ 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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