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Problem:

$F = \frac{(G m_1 m_2)}{r^2}$

An object of unknown mass is placed directly between the Earth and the Moon such that the forces of gravity acting on the object are equal from both the Earth and the Moon. The distance from the Moon to the object is exactly one-tenth of the distance from the Moon to Earth. Using the above formula, determine the distance from the Moon to the Earth.

Given is:

The factor $G≈6.67∙10−11\mathbf{N(\frac{m^2}{kg^2})}$ is a universal constant.

Mass of Earth $≈5.972∙10^{24}\mathbf{kg}$

Mass of the Moon as $≈7.348∙10^{22}\mathbf{kg}$

Mass of unknown object = $m$

My solution:

$F = \frac{(6.67 *10^-11) m (7.348*10^{24})}{R^2}$

$F = \frac{(6.67 *10^-11) (5.92* 10^22) m}{10R^2}$

As gravitational force acting on the object is equal

$\frac{(5.92\cdot 10^22)}{100r^2} = \frac{(7.348*10^{24})}{r^2}$

$R = 1.24 \cdot 10^4 KM$

Is this a viable solution to the problem? Please provide feedback. Thank you!

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  • $\begingroup$ First, the answer is wrong. The distance from the earth to the moon is about $384,400$ km. Next, you've used the wrong powers of ten in your equations for $F$. Third, you should be using $9R$, not $10R$. But most importantly, I don't think there's enough information to answer the question. The ratio of the two masses is (to two significant digits) $81$, so the distance could be anything. $\endgroup$
    – rogerl
    Nov 14, 2016 at 0:35
  • $\begingroup$ Yeah that is actually what I thought. Me and my friend are doing this question and he asked me to check if he was right. Funny situation. Thank you for your input. $\endgroup$
    – Arthur Alex Karapetov
    Nov 14, 2016 at 1:26

1 Answer 1

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While I haven't checked that you got the right numbers, I would make the following comments:

The approach looks correct. However, you need to put some more context around your equations. Don't just write "F = blah, F = blah, therefore blah = blah". Set out that $r$ is the distance from x to y, then say that the gravitational force due to the moon is $F_1$, and due to the Earth is $F_2$, and that they're equal so therefore blah.

The more context and information you provide, the easier it is to see how much you actually understood of the problem, and the easier it is to spot where you might have made a mistake, which lets the marker be more generous about applying partial marks (and it makes it easier for you to spot the mistakes and correct them before you hand the homework in).

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  • $\begingroup$ What you said is noted. Thank you very much for your input. $\endgroup$
    – Arthur Alex Karapetov
    Nov 14, 2016 at 1:27

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