<p>Consider a particle with mass m, orbiting in a circle  a body with mass M.  The gravitational force must be the centripetal force causing circular motion, so</p>
<p><span class="math-container">$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$</span></p>
<p>Cancelling the <span class="math-container">$m$</span> and <span class="math-container">$r$</span> and square rooting gives:</p>
<p><span class="math-container">$$ \frac{\sqrt{GM}}{\sqrt{r}} = v$$</span></p>
<p>So you see the velocity is inversely proportional to the square root of distance <span class="math-container">$r$</span>.  If the distance doubles, the velocity decreases by a factor of <span class="math-container">$1/\sqrt{2}\approx 0.707$</span>.  This is the 70.7% that you observed.</p>

CC BY-SA 4.0
2021-11-11T22:44:07.573Z
2021-11-11T22:44:07.573Z
James K
<p>Looking at the speeds of the planets in our solar system when traveling around the sun and their distances from it (or, using Kepler's third law,), it seems that when a planets distance is doubled, it's velocity drops to about 70.7% of its previous value...</p>
<p>Why is that, if the gravitational strength between the planet and sun is now only a quarter of its previous value?  Shouldn't it be moving at 25% the speed it would have if it was half as distant, not 70.7%?...</p>
<p>All I can think of is that the more distant planet doesn't have to expend as much 'gravitational energy' or whatever changing its direction (which is a form of 'acceleration'), because it is moving in a straighter line (less curvature; less angle in its angular velocity) during its orbit....</p>

<orbit><planet><gravity><orbital-mechanics><johannes-kepler>
Why do planets' orbital velocities drop off by less than a third when twice as distant from the sun, rather than three-quarters? If gravity is $1/4$?
Kurt Hikes
# Why do planets' orbital velocities drop off by less than a third when twice as distant from the sun, rather than three-quarters? If gravity is $1/4$?

(Asked by: username_0 on 2021-11-11)

Looking at the speeds of the planets in our solar system when traveling around the sun and their distances from it (or, using Kepler's third law,), it seems that when a planets distance is doubled, it's velocity drops to about 70.7% of its previous value...

Why is that, if the gravitational strength between the planet and sun is now only a quarter of its previous value? Shouldn't it be moving at 25% the speed it would have if it was half as distant, not 70.7%?...

All I can think of is that the more distant planet doesn't have to expend as much 'gravitational energy' or whatever changing its direction (which is a form of 'acceleration'), because it is moving in a straighter line (less curvature; less angle in its angular velocity) during its orbit....

**Tags:** orbit, planet, gravity, orbital-mechanics, johannes-kepler


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## 1 answers

### Answer by username_1 on 2021-11-11. Score: 7 | Accepted answer

Consider a particle with mass m, orbiting in a circle a body with mass M. The gravitational force must be the centripetal force causing circular motion, so

$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$

Cancelling the $m$ and $r$ and square rooting gives:

$$ \frac{\sqrt{GM}}{\sqrt{r}} = v$$

So you see the velocity is inversely proportional to the square root of distance $r$. If the distance doubles, the velocity decreases by a factor of $1/\sqrt{2}\approx 0.707$. This is the 70.7% that you observed.


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