<p>Using this equation from <a href="https://en.m.wikipedia.org/wiki/Gravitational_wave#Binaries" rel="nofollow noreferrer">Wikipedia</a> for the approximate merger time:</p>
<p><span class="math-container">$$\frac{5}{256}\frac{c^5}{G^3}\frac{r^4}{m_1m_2(m_1+m_2)},$$</span></p>
<p>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:</p>
<p><code>((5/256) * (c^5/G^3) * ((100 miles) + (1738.1 km))^4 / ((11110 kg) * (7.347631E22 kg)^2))</code></p>
<p>to perform the calculation.</p>
<p>The result is <span class="math-container">$\approx 3.4497 \times 10^{46}$</span> seconds, or around <span class="math-container">$1.093 \times 10^{39}$</span> years, which is about <span class="math-container">$7.93 \times 10^{28}$</span> times the current age of the universe.</p>
<p>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.</p>
<p>There's some information about the effect of orbit eccentricity  on changes to orbital angular momentum and energy due to gravitational radiation at <a href="https://en.wikipedia.org/wiki/Two-body_problem_in_general_relativity#Gravitational_radiation" rel="nofollow noreferrer">Two-body problem in general relativity</a>.</p>

CC BY-SA 4.0
2021-02-21T15:43:40.040Z
How off will this estimate be when the satellite approaches very close to the surface of the object? (like the final 1 mi, etc)
slowerthanstopped
CC BY-SA 4.0
2021-02-21T15:51:11.500Z
@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" $$r(t)=r_0(1-\frac{t}{t_{coalesce}})^{\frac14}$$
PM 2Ring
CC BY-SA 4.0
2021-02-21T15:55:52.593Z
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.
PM 2Ring
CC BY-SA 4.0
2021-02-21T15:24:28.417Z
2021-02-21T15:24:28.417Z
PM 2Ring
<p>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.</p>
<p>How long would it take for the satellite to crash into Body A via gravitational emission?</p>

<gravity><gravitational-waves><satellite><hypothetical><orbital-decay>
Orbital decay via gravitational emission in a hypothetical, two-body universe
slowerthanstopped
# Orbital decay via gravitational emission in a hypothetical, two-body universe

(Asked by: username_0 on 2021-02-20)

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?

**Tags:** gravity, gravitational-waves, satellite, hypothetical, orbital-decay

---

## 1 answers

### Answer by username_1 on 2021-02-21. Score: 1 | Accepted answer

Using this equation from [Wikipedia](https://en.m.wikipedia.org/wiki/Gravitational_wave#Binaries) 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](https://en.wikipedia.org/wiki/Two-body_problem_in_general_relativity#Gravitational_radiation).

> **Comments:**
> **username_0**: How off will this estimate be when the satellite approaches very close to the surface of the object? (like the final 1 mi, etc)  
> **username_1**: @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" $$r(t)=r_0(1-\frac{t}{t_{coalesce}})^{\frac14}$$  
> **username_1**: 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.  

---

