# Decelerating satellite

1. Jun 5, 2010

### Bjarne

Let’s assume a satellites orbit a area with resistance against motion, for example 1E-10m^2

If that motion would be straight an object would deceleration, but a satellite would instead (in its accelerating period) continue to accelerate, also even it had such speed lost.

On the one had the satellite would lose 1E-10m^2 per second.
But on the other hand it would also (in its accelerating period) gain more speed because now it is falling faster to the Earth.

To calculate the loss of speed should be easy = ( t * 1E-10m^2)
But how can I calculate how much speed the satellite gain due to now it also will fall a bit faster to the earth?

(Let say the satellite orbit at a radius (from the center of the earth) = 35000 km.)

(This is not home work)

2. Jun 5, 2010

### Janus

Staff Emeritus
Let's see, an orbit at 35,000 km has a period of 112,617 sec and a orbital velocity of 3382.222431 m/s.

So, over the course of one orbit, it will "lose" 1.3e-5 m/s.

The energy of the satellite is found by

$$E = \frac{mv^2}{2}- \frac{GMm}{r}$$
Where r is the length of the radial vector.
or

$$E = -\frac{GMm}{2a}$$

Where 'a' is the average orbital distance.

By solving the first equation with the new orbital velocity after one orbit (3382.222431-1.3e-5 m/s) , equating that to the second equation and solving for a, we get the new average orbital distance.

Assuming that this new orbit is still circular, we can use

$$v = \sqrt{\frac{Gm}{a}}$$ to find the new orbital velocity.

This gives a new orbital velocity of 3382.222444 m/s after one orbit (and a new orbital distance of 34999.99973 km
.

However, this is only an estimate, as it takes some short cuts. For example, for ease of calculation, it is assumed that the total velocity is lost at one instant instead of slowly over the whole orbit, and that the new orbit will be circular.

3. Apr 27, 2011

### Bjarne

Agree.
[PLAIN]https://www.physicsforums.com/latex_images/27/2748362-2.png [Broken] [Broken]
Let us just say; - sqrt(6.67e-11*5.97e24/35000000.000)= 3372.997822041001 m/s
(This make it easier for me to compare later, I don't know the data you have used. )

I Disagree.
If the orbit (measured from the center of the Earth) = (average)= 35,000km, - then the orbit-time; = 35,000,000*3,14*2/3372.997822041001 m/s = 65165 s. – (one complete orbit).

I Disagree.
We don’t know the speed-change (increase) at this point, but can only calculate the distance lost, so far.
Actual there is no direct speed lost (in the acceleration period) because deceleration is "converted" to acceleration in this case.
So in this first steep we can only calculate the distance change.
In one complete orbit the satellite would have a distance-lost

Average distance loss per/s. ; = 65165 s.* 1e-10 m/s /2 = 3.25825e-6 m/s
Total Distance lost; 65165s * 3.25825e-6 = 0.21232386125 Meter

[PLAIN]https://www.physicsforums.com/latex_images/27/2748362-2.png [Broken] [Broken]

Now we only need to enter the new radius to compare the orbital average speed..

sqrt(6.67e-11*5.97e24/34999999.78767614)= 3372.997829043529 m/s
sqrt(6.67e-11*5.97e24/35000000.00000000)= 3372.997818812557 m/s

Speed difference .................................= 1.023097183860955e-5 m/s

Yes, right the result I just found is wrong, - because the basic deceleration taking place during 1 orbit will also affect the speed of the "the new orbit"
The acceleration due to approaching the Earth will therefore not accelerate so much in that period.

[PLAIN]https://www.physicsforums.com/latex_images/27/2748362-0.png [Broken]

I don’t think energy change equation is usefull in this case, because then we must know the reel new speed first..?

So - how can the real speed be calculated, after the resistance have been going on during 1 complete orbit..

Which equation can be used to know calculate the acceleration rate from apogee towards perigee? -
(I think may this can be used somehow as the last one)

Last edited by a moderator: May 5, 2017
4. Apr 28, 2011

### IsometricPion

If the initial orbit is circular and the acceleration is a constant 1E-10 m/2 in the direction opposite that of the instantaneous velocity at any given time: $$r=\frac{GM}{(\sqrt{\frac{GM}{r_{o}}}-\alpha{}t)^2}$$ where $$\alpha$$ is the constant acceleration's value and M is Earth's mass.

For t=65165, the new distance is about 0.13517 m greater than the initial value, ro. The tangential velocity is 6.5165E-6 m/s less than its initial value.

Note that a constant tangential acceleration is not what would normally be expected due to air resistance or any other natural "resistance" of which I am aware.

Last edited: Apr 28, 2011
5. Apr 28, 2011

### Bjarne

You mean the new distance is about 0.13517 m less ?

What went wrong here?

[PLAIN]https://www.physicsforums.com/latex_images/32/3272376-0.png [Broken]

6.67e-11*5.97e24/sqrt(6.67e-11*5.97e24-1e-10/35000000)^2

I get, - 1

Let's assume deceleration from aphelion and towards perihelion.
What about the new perihelion speed?

Last edited by a moderator: May 5, 2017
6. Apr 28, 2011

### IsometricPion

If you calculated it exactly as you wrote here, you entered the wrong expression it should have been: 6.67e-11*5.97e24/(sqrt(6.67e-11*5.97e24/35000000)-1e-10*65165)^2. When one exerts a force in order to decelerate and object, energy is added to the system. In the case of gravity, this causes the separation between the two bodies to increase (since one is decreasing both the angular momentum and binding energy of the system). So, it is expected that the new perihelion speed is smaller than the old and that the new perihelion is at a greater radius than the old.

7. Apr 29, 2011

### Bjarne

8. Apr 29, 2011

### D H

Staff Emeritus
That makes *no* sense. Think about it for a bit. Drag is directed against the velocity vector, more or less, so drag is a good exemplar of what Bjarne was asking about in the opening post. (To be pedantic, drag is directed against velocity relative to that of the atmosphere, rather than the orbital velocity. However that velocity to the wind is very close to the orbital velocity vector and the orientations of the wind-relative velocity and the orbital velocity are very, very close to one another.)

A spacecraft subject to drag does not spiral out. It spirals in.

Seemingly paradoxically, a spacecraft subject to a small amount of drag increases in speed as it spirals in. At the point where drag does make a spacecraft slow down rather than speed up, the drag is no longer just a small perturbing force. It is now a very significant force. This is essentially the point at which the spacecraft can be said to have re-entered the atmosphere.

Yeah, but it's a pretty good approximation, at least for quite a few orbits.

Last edited: Apr 29, 2011
9. May 6, 2011

### IsometricPion

That makes sense. I guess I was using my intuition in an arena in which it did not apply. Should have known an analytic solution was too good to be true.

So, I took the equations of motion of the two body problem in center-of mass coordinates: $$2\dot{r}\dot{\theta}+r\ddot{\theta}=0, \ddot{r}=r\dot{\theta}^{2}-\frac{k}{\mu{}r^2}\text{ where }\mu{}=\frac{m_{Earth}m_{satellite}}{m_{Earth}+m_{satellite}}\text{ and }k=Gm_{Earth}m_{satellite}$$ and replaced the zero on the right-hand side of the first equation (which, I assume, corresponds to the acceleration in the θ-direction) with -1E-10. Numerically integrating these equations I obtain that after 65165s, rinital-rfinal≈0.13517m and the tangential speed (v=r$$\dot{\theta}$$) difference is vfinal-vinitial≈6.5164e-6m/s. The initial conditions were those of a circular orbit of radius 3.5E7m. These new values seem to be in accord with everyone's expectations in sign and order of magnitude. Though, I don't think the values are exactly correct since in the center-of-mass coordinates what I did probably corresponds to giving approximately correct value of acceleration to the satellite and a very small amount of acceleration to Earth as well (I think the value would be many orders of magnitude smaller than the acceleration on the satellite.). Any satellite mass below of order 1E20kg will give approximately the same values as I gave above for the time specified. The uncertainty in the values given above is dominated by the uncertainty in G and Earth's mass (I used G=http://physics.nist.gov/cgi-bin/cuu/Value?bg|search_for=universal_in!".).

Last edited by a moderator: Apr 25, 2017
10. Mar 15, 2012

### Bjarne

Thank you all so far.
Let’s say that the satellite was moving (from apogee) towards perigee and exactly from the perigee direction there is a very strong solar wind.

The solar wind would resists the motion with let’s say exactly 1E-9 m/s
After passing perigee the opposite happens, now the e solar wind would be “following wind” and “push” the satellite 1E-9 m/s faster towards apogee.

This would force the orbit more elongated, but I guess it also would cause its own Perigee Precession Anomaly.

How can the Perigee Precession Anomalybe be calculated.

Last edited: Mar 15, 2012
11. Mar 20, 2012

### IsometricPion

I went back and reworked the math for the problem you originally posted. I came up with some new equations of motion that seem to give somewhat smaller changes over the course of an orbit (ri-rf~6.87E-2m vf-vi~1.07E-7m/s). A similar analysis qualitatively confirms the idea that an acceleration that switches sign at periapsis produces an increase in the eccentricity of the orbit. I have not yet done any analysis on the apsidal precession, but I would not expect any if the direction of the "wind" remained the same (which wouldn't be true if the wind is provided by a star which the planet (and satellite) is orbiting).