# At what velocity will the moon be able to escape the earth's orbit?

I haven't found an answer to this question online so correct me if I am wrong.

This is the escape velocity equation:

$$v_{esc} = \sqrt{\frac{2GM}{r}}$$

and by plugging the known values, this is the velocity needed for the moon to escape earth's orbit:

$$v_{esc} = \sqrt{\frac{2 \times 6.674 \times 10^{−11} \times 7.348 \times 10^{22}}{384402}} \approx 5051.26$$ km/s



Did I do anything wrong?

• We don't check calculation results on this site. But you need to work in metres, you can convert to km/s after you take the square root. Also, $M$ should be the mass of the Earth, not the Moon (although that's only accurate for bodies much lighter than the Earth). – PM 2Ring Feb 19 '19 at 15:33
• See physics.stackexchange.com/a/121834/123208 for the derivation of a more accurate formula. – PM 2Ring Feb 19 '19 at 15:59
• Thank you for the feedback. About what you said inside the brackets; if I were to calculate the escape velocity of the ISS from earth's orbit, would the answer be more accurate since the ISS has a significantly lighter mass than the earth? Edit: Nevermind, I got it – George V Feb 19 '19 at 16:00
• Yes, your formula is fine for the ISS. – PM 2Ring Feb 19 '19 at 16:03
• I just edited my comment the moment you replied. Thank you :) – George V Feb 19 '19 at 16:04

I now understood the topic much better thanks to @PM 2Ring's comments. The mass of the moon is too heavy to be neglected so the equation for the escape velocity needs to be written like this:

$$v_{esc} = \sqrt{\frac{2G(M+m)}{r}}$$

where:

M = mass of the Earth

m = mass of the Moon

$$v_{esc} = \sqrt{\frac{2 \times 6.674 \times 10^{−11} (5.972 \times 10^{24} + 7.348 \times 10^{22})}{3.84402 \times 10^{8}}} = 1448.87$$ m/s