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I used the equation F = Gm1m2/d2. I'm trying to find the distance (d). I have G (7.33 m/s^2). But I also don't have F or m1 and m2. How do I find these three unknown variables?

Dylan
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  • You would have luck asking in Physics.se – JavaMan Oct 01 '19 at 01:08
  • Well the gravitational force is inversely proportional to the square of the distance (from the center of the Earth and acceleration is proportional to force. You can look up the acceleration at the surface and the radius of the Earth. Unless I've misremembered physics that should be enough information. – Ethan Bolker Oct 01 '19 at 01:15

2 Answers2

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You want the acceleration due to gravity, so the test mass is irrelevant. What you know is that $$GM/R^2 = g,$$ where $M$ is the mass of the earth, $R$ is the radius of the earth, and $g \approx 9.8 \text{m/sec}^2$, as this gives the acceleration at the surface of the earth.

You want to solve $GM/r^2 = 7.33$, knowing that $GM/R^2 = 9.8$. Can you do this algebra? Remember, when you've done this, that you want the distance from the surface of the earth, so you want $r-R$.

Ted Shifrin
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  • In fact, somewhere inside the earth, the acceleration due to gravity is also $7.33m/s^2$... can you cover this situation? – WhatsUp Oct 01 '19 at 01:28
  • Yes, but the formula is different inside the earth. Inside the earth, the acceleration is a linear function of $r$ (amazingly!). So you would need to solve $gr/R = 7.33$. To show this formula is more advanced (although Newton figured it out) :) – Ted Shifrin Oct 01 '19 at 01:32
  • Wait... you assumed that the density is uniform inside the earth?! :P – WhatsUp Oct 01 '19 at 01:35
  • Fair point! As @WhatsUp points out, you need to assume that the earth is perfectly round and of constant density to deduce the formula I just gave. :) – Ted Shifrin Oct 01 '19 at 01:36
  • Ja, it's a small trap... :) The original answer is fine for outside the earth. – WhatsUp Oct 01 '19 at 01:39
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At the surface of the earth, you have

$$mg=\frac{Gm M}{R^2}\tag{1}$$

where $g=9.8m/s^2$ and $R$ is the radius of the earth. Similarly, at the distance $h$ from the surface,

$$mg'=\frac{Gm M}{(R+h)^2}\tag{2}$$

where $g'=7.33m/s^2$. Take the ratio of (1) and (2),

$$\frac{g'}{g} = \frac{R^2}{(R+h)^2}$$

Then, the distance $h$ is given by

$$h=R\left(\sqrt{\frac {g}{g'}}-1 \right)$$

Quanto
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