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Assume we have a universe with nothing in it except for two objects. One of them is a stationary object that is perfectly spherical with the mass of the Moon (without any atmosphere, of course; also, let's call this Body A). The other body is a small satellite similar to the Hubble Space Telescope, orbiting at 100 miles (161 km) above the surface of Body A.

How long would it take for the satellite to crash into Body A via gravitational emission?

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Using this equation from Wikipedia for the approximate merger time:

$$\frac{5}{256}\frac{c^5}{G^3}\frac{r^4}{m_1m_2(m_1+m_2)},$$

and Wikipedia's values for the Moon's equatorial radius and mass, and the launch mass of the HST, we can use this query in Google:

((5/256) * (c^5/G^3) * ((100 miles) + (1738.1 km))^4 / ((11110 kg) * (7.347631E22 kg)^2))

to perform the calculation.

The result is $\approx 3.4497 \times 10^{46}$ seconds, or around $1.093 \times 10^{39}$ years, which is about $7.93 \times 10^{28}$ times the current age of the universe.

That equation is for a circular orbit, and it's only a crude estimate based on orbital energy lost via gravitational radiation. It mostly ignores the change in the orbital radius, since that's fairly small until the final stages of the merger.

There's some information about the effect of orbit eccentricity on changes to orbital angular momentum and energy due to gravitational radiation at Two-body problem in general relativity.

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  • $\begingroup$ How off will this estimate be when the satellite approaches very close to the surface of the object? (like the final 1 mi, etc) $\endgroup$
    – slowerthanstopped
    Feb 21, 2021 at 15:43
  • $\begingroup$ @slowerthanstopped See the 1st link, just before the equation I quoted. "This estimate overlooks the decrease in r over time, but the radius varies only slowly for most of the time and plunges at later stages" $$r(t)=r_0(1-\frac{t}{t_{coalesce}})^{\frac14}$$ $\endgroup$
    – PM 2Ring
    Feb 21, 2021 at 15:51
  • $\begingroup$ In the real universe, this calculation is mostly irrelevant because it ignores friction due to matter in space, forces from electromagnetic radiation, changes in the gravitational environment, etc. But I guess it's ok as a first approximation. $\endgroup$
    – PM 2Ring
    Feb 21, 2021 at 15:55

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