# Soyuz/spacecraft deceleration speed for reentry

I recently saw a program by Prof. Brian Cox (Human Universe Ep.1) where he mentioned that just by using two equations - f=ma and the universal law of gravitation, you could calculate how much a spacecraft would have to slow down by to enter a spiral decent for a reentry. With orbital velocity known, he stated that the astronauts would need to slow down by 128 m/s and gravity would do the rest.

I've tried to figure how this could be, if there is anything more than just employing two formulae.

• That 128 m/s slow down speed is not valid for all orbit heights, only for low orbits.
– Uwe
Sep 29 '20 at 19:11
• The deorbit burn serves only to dip the perigee into the sensible atmosphere. Drag does the rest. See space.stackexchange.com/q/12011/6944, which this question may be a duplicate of. Sep 30 '20 at 0:59
• Are these two formulas enough? Yes. The caveat is you should solve the differential equation F=ma, and this task is beyond high school math (usually). Actually a (acceleration) is second derivative of coordinate vector r, and F is gravitational force that depends on r. If we solve the equation we'll see that trajectory of a spacecraft is ellipse (if velocity is below escape velocity). If a spacecraft needs to land on Earth, it should lower its speed, so the lowest point of its elliptic orbit reaches upper atmosphere (about 100 km above surface). Atmospheric drag will do the rest. Sep 30 '20 at 9:38
• ...But if we already know the trajectory is ellipse we can simplify the task by using energy conservation principle. While a spacecraft moves beyond atmosphere and don't fire its engines - spacecraft's total energy will be constant. This principle is used in "vis viva equation". You can read about the term in Wikipedia, as well as about "Hohmann transfer" and other interlinked themes in the Wikipedia articles. I hope it can help. Sep 30 '20 at 9:50
• "to enter a spiral decent for a reentry" is possible only with the atmospheric drag. No drag, no spiral. So you need to know not only the orbit but also a height were spiraling starts, Calculation of the slow down speed is impossible without that height.
– Uwe
Oct 7 '20 at 8:51

It's true that you could calculate how much a spacecraft would have to slow down by to enter a spiral decent for a reentry from frist principles.

A "spiral" descent isn't an effect of gravity alone (things don't orbit in spirals), but an rather an effect of atmospheric drag.

As such, the part to calculate is really "how do we get low enough to get inside the atmosphere?". After getting inside the atmosphere, drag (and gravity) will do the rest with no further thrust needed.

Here's what this idea looks like: All that is needed for this calculation is the following formula:

$$v = GM\sqrt{\frac{2}{r} - \frac{1}{a}}$$

You can see the entire 12-step derivation of this equation at Wikipedia, the vis-viva equation It uses nothing but the law of gravity, conservation of energy and momentum, and geometry. That's exactly as close to the metal as what Brian Cox claims.

Using it is very simple. See that part to the right in the diagram where the black (original) orbit and red (desired) orbit touch each other? We only need the difference in velocity at that point.

The vis-viva equation tells us exactly what the velocity ($$v$$) is at any part of an orbit, using only four numbers we already have available:

• $$G$$, the universal gravitational constant, from the law of gravity.
• $$M$$, the mass of the Earth.
• $$r$$, the current orbital radius.
• $$a$$, the semi-major axis of the orbit.

The semi-major axis is simply the average of the lowest point and highest point of the orbit.

For the first orbit, $$a = r$$ since it's circular. For the second orbit, $$a$$ is the average of $$r$$ and the distance between the top of the atmosphere and the centre of the Earth.

You then have two values for $$v$$, which you can subtract from each other to reach the same conclusion as Brian Cox.

• My answer already demonstrated the necessity of the vis-viva equation and asserted that since conservation of energy is not mentioned by Cox, Cox is wrong. This answer basically reiterates what I wrote except that it tries to support the opposite conclusion. I strongly disagree with "That's exactly as close to the metal as what Brian Cox claims." No it is not. He left out conservation of energy. Cox is wrong.
– uhoh
Oct 6 '20 at 22:26
• Cox's words (from my answer): "All you need are the two laws, written down first by Isaac Newton: $$F = ma$$ and the universal law of gravitation: $$F = \frac{GmM}{r^2}. \text{"}$$ Cox said nothing about conservation of energy, so I feel that his assertion is wrong, as is this answer.
– uhoh
Oct 6 '20 at 22:46
• -1 for all of the above, but +1 for combining purple and cyan in a graphic, and for a very nice and quite instructive discourse
– uhoh
Oct 6 '20 at 22:53
• Conservation of energy is an uncontroversial a-priori assumption. Oct 6 '20 at 23:06
• Why would $$\frac{v^2}{2} - \frac{GM}{r} = \text{const}$$ be "an uncontroversial a priori assumption" while at the same time $$F=ma$$ had to be written out explicitly? Your argument is not consistent with Cox's "all you need" thesis. Cox is wrong. And that's the answer to the OP's question as asked: "I've tried to figure how this could be, if there is anything more than just employing two formulae."
– uhoh
Oct 6 '20 at 23:09

### Maybe Cox is wrong?

What was said?

I found currently viewable copies on YouTube and Daily Motion and transcribed a bit after about 40:00 as follows:

All you need are the two laws, written down first by Isaac Newton:

$$F = ma$$

and the universal law of gravitation:

$$F = \frac{GmM}{r^2}$$.

Now what you can show from those, really simply, is that for a circular orbit, which is what the International Space Station is basically in, the velocity (flying along there) is given by

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

where M is the mass of the Earth and r is the distance to the center of the Earth.

The explanation goes on, but the third equation, the "Now what you can show from those, really simply, is that..." bit is a form of the vis-viva Equation

$$v = \sqrt{GM \left( \frac{2}{r} - \frac{1}{a} \right)}$$

but simplified for a circular orbit where $$r=a$$.

The derivation of the vis-viva Equation is not short, and generally requires the conservation of energy, and the understanding that it is the sum of kinetic and potential energies:

$$E = T + P = \frac{1}{2}v^2 - \frac{GM}{r} = \text{const}$$

and these are reduced energies, the mass of the object is dropped because it divides out.

Using integration we can get

$$P = -\frac{GM}{r}$$

from

$$F = \frac{GmM}{r^2}$$

by integration and paying attention to signs.

But I don't see how we can get

$$T = \frac{1}{2}v^2.$$

## It's possible he was wrong.

I say that because there's another mistake on the same page! He works out the orbital velocity for the ISS to be 7358 meters per second.

It would be if the ISS were up around 1000 km of altitude, but it's not. At 400 km the ISS' velocity is closer to 7670 meters per second.

Of course one can argue that Cox used a spherical cow/horse and that the ISS is closer to 1000 km than 100 km on a logarithmic axis, but I think (though I'm not sure) that he might not be right and that we need one more equation. • While the derivation of vis-viva equation is not short, it is easy to get the speed of a circular orbit if you use those two formulas and the formula for acceleration of a movement along a circle with a constant speed: $a=v^2/r$. Oct 1 '20 at 7:38
• For a circular orbit, acceleration has to be equal to $v^2/r$. From those formulas, the acceleration is equal to $GM/r^2$. From the equality of these two expressions, we get the formula for $v$. Oct 7 '20 at 8:27
• Concerning your "extended rant": one possible interpretation of Cox's words (I haven't seen the video, so this interpretation may be too generous) is: those two are the only physical laws we need to know to derive the vis-viva equation etc. Everything else we need is just mathematics. This mathematics may be non-trivial, but it is something which can be deduced purely intellectually, without making any experiments. Oct 7 '20 at 8:33
• Well, it's a bit more tricky. I think that to derive the vis-viva equation, you also need the conservation of angular momentum, but again, it follows from $F=ma$ and the fact that the force is radial. Oct 7 '20 at 9:07
• For circular orbits, sure. But you need the vis-viva equation for non-circular orbits to know how much you need to decrease speed to lower the periapse into atmosphere. (If you know about the conservation of energy and that the reduced energy of an orbit is equal to $-GM/2a$, then the vis-viva equation follows from that. But to deduce that the energy depends only on the semi-major axis like this, you also need the conservation of angular momentum.) Oct 7 '20 at 10:06