# Satellite Collision Problem

I'm stuck on a problem of two satellites going towards each other. The mass of the first satellite is 400kg and the mass of the second satellite is 100kg. The altitude of the satellites is 1000km. I want to know if once the satellites collide, will they continue in orbit or will they crash and burn into the Earth?

I know that you can get the velocity of each satellite by using this equation:

$$v = \sqrt{GM/R}$$

Where $v$ is the velocity, $G$ is $6.67\cdot 10^{-11}$, $M$ is the mass of the planet, and $R$ is the radius.
I can get the velocities, but i don't know where to go from here. Also when I get $R$, do I also have to add the radius of the planet as well? so it would be 1000km + radius of planet?

If you guys would help me with this, it would be amazing.

• Should this be tagged "homework"? Sep 3, 2014 at 7:31
• @Floris: it's a close call. I think there is an interesting concept here, i.e. the application of the vis viva equation, that might be interesting to others. Sep 3, 2014 at 7:43
• I'm OK to answer interesting questions, and agree this is one - just thought the tag might be appropriate. Really I'm asking OP... Sep 3, 2014 at 7:54

Unless I'm missing an easy way to do this problem it seems a surprisingly hard one. This diagram shows the problem (I've exaggerated the altitude of the satellite to make the diagram clearer):

The satellites are in circular orbits (dotted line) at a distance $r$ from the centre of the Earth, so their orbital velocity as (as you say):

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

Assuming the two satellites stick to each other when they collide you end up with a single 500kg mass of twisted metal moving at some velocity $v'$ that is less than $v$. The new orbit will be an ellipse, and the question is whether the new orbit intersects the Earth's surface. If the new orbit does not intersect the Earth then the fused satellites will continue to orbit, while if the new orbit does intersect the Earth obviously the debris will crash into the Earth.

The first step is to calculate v', and this is done simply by conservation of momentum. At the moment before the collision the momentum of the 400kg satellite is $400v$ and the momentum of the 100kg satellite is $-100v$ (I'm taking velocity positive to the right). After the collision the momentum of the wreckage is $500v'$, so:

$$400v - 100v = 500 v'$$

and we get:

$$v' = \frac{3}{5}v \tag{1}$$

The hard bit is working out the new orbit. For this you need to start with the vis viva equation, and with some head scratching you can work out the equation for the perigee distance ($r_p$ in the diagram):

$$r_p = \frac{r_a}{\frac{2GM}{v'^2r_a} - 1}$$

The question is then simply whether $r_p$ is less than the radius of the Earth.

You know $v'$ from equation (1), and $r_a$ is just the initial orbital radius (measured from the centre of the Earth). I'll leave it up to you you calculate $r_p$ and answer the question.

• There is an easy solution - using the fact that orbital energy relates to major axis in a very simple way. Can't write out the math on my phone but you start with KE=PE for circular orbit, reduce KE to 9/25th, then find the major axis and see your distance from the center at apogee is 9/25 of distance at perigee. Sep 3, 2014 at 7:28
• @Floris - Easy, but wrong. John's answer gives a result considerably smaller than yours. Sep 3, 2014 at 21:45
• @JohnRennie - Your answer simplifies considerably if you recognize that $GM/r_a$ is $v^2$. Another way to arrive at the same numerical result is to use the concept of the eccentricity vector to compute the eccentricity $e = 1-\frac{v^2 r}{GM}$ at periapsis or apoapsis and then use the fact that $\frac{r_a}{r_p} = \frac {1+e}{1-e}$. Sep 3, 2014 at 21:48
• @DavidHammen - you are right; I had lost a factor 2 along the way (put KE=PE, instead of 2KE=PE). That's what happens when you do physics on a napkin in an airport with jetlag. I have now written out my approach step by step, and get the same answer as John. Sep 3, 2014 at 23:43
• @DavidHammen your equation for the eccentricity is true at apoapsis, however if you would take the absolute value then it would also be true at periapsis. It general you can find it with $e=\sqrt{1+\frac{v_{\theta}^2r}{GM}\left(\frac{v^2r}{GM}-2\right)}$, which at periapsis or apoapsis is equal to $e=\sqrt{\left(1-\frac{v^2r}{GM}\right)^2}$. Sep 4, 2014 at 0:02

Introducing a side issue to this question:

Using $v$ for the orbital velocity at $1000$ km altitude, the total kinetic energy of the two satellites just before the collision is $$KE_{\text{Before}}=\frac12400v^2+\frac12100v^2=250v^2$$and using the final velocity of John Rennie above, the total kinetic energy of the combined satellites just after the collision is :$$KE_{\text{After}}=\frac12500\left(\frac35v\right)^2=90v^2$$The $160 v^2$ joules of lost energy is distributed as heat over the $500 \text{ kg}$ satellite. Thus, each kilogram will absorb $0.320v^2$ joules of energy.

The orbital velocity at an altitude of $1000\text{ km}$ above the surface of the earth is $7.35\text{ km/sec}$, so each kilogram of combined satellite will need to absorb 17.27 Mega joules of energy.

If we assume that the two satellites are made of aluminum, this implies a temperature increase of around $19, 000 \text{ Kelvin}$. Even if we assume that the two satellites have the same high specific heat as water, we get a temperature increase of over $4, 100 \text{ Kelvin}$

We have to make a few assumptions here, because your question as posed is a bit incomplete.

1) collision is inelastic
2) satellites don't disintegrate on collision (see user58220's interesting observation)
3) there is no atmosphere (so drag is not an issue - let's see if the satellites end up in an orbit above the planet's surface)
4) you said "crash into the earth" so I assume R = 6400 km. The value of R does matter

After the collision, the combined velocity of the two satellites (with initial velocity $v$) will be given by conservation of momentum:

$$(m_1 + m_2) v_{final} = m_1 v_{initial} - m_2 v_{initial}\\ v_{final} = v_{initial}\frac{300}{500}$$

You now have satellites at the apogee of an elliptical orbit, and they will approach the earth until their closest point, the perigee. Let us say that the radial distance at apogee was $R+h_a$, and at perigee $R + h_p$.

From orbital mechanics we know there is a relationship between the semimajor axis of the orbit and the orbital energy:

$$-\frac{\mu}{2a}=\epsilon\tag1$$

We know the orbital energy decreased by the collision: the kinetic energy before was

$$KE=\frac12(m_1 + m_2)v^2$$

and afterwards it is

$$KE = \frac12\left(\frac{3}{5}\right)^2(m_1 + m_2)v^2$$

In circular orbit, the kinetic energy is equal to minus half the potential energy; after the collision, the potential energy is unchanged (at the instant of collision) while the kinetic energy was decreased. So

\begin{align}\epsilon_{before}&=\frac{v^2}{2}-\frac{\mu}{r}\\ &= KE_{before} - 2 KE_{before}\\ &= -KE_{before}\end{align}\tag2

The kinetic energy afterwards is $\frac{9}{25}^{th}$ of the kinetic energy before, so we can write

\begin{align}\\ \epsilon_{after} &= (\frac{9}{25}-2)KE_{before}\\ &= -\frac{41}{25}KE_{before}\\ \end{align}\tag3

Combining equations (1), (2) and (3) gets me the ratio of the major axes:

\begin{align}\\ \frac{a_{before}}{a_{after}}&=\frac{\epsilon_{after}}{\epsilon_{before}}\\ &=\frac{25}{41}\\ \end{align}

For the circular orbit we can say that $a=2r_a$, while for the elliptical orbit it is $r_a+r_p$. So now we have

$$\frac{2r_a}{r_a+r_p}=\frac{41}{25}$$

With a bit of rearranging this gives

$$r_p = \frac{9}{41}r_a$$

So when the radius of the earth is about 6400 km, and you are orbiting 1000 km above the surface, you will surely crash. The largest planet on which this collision would not result in termination of the orbit has a radius of

$$r_{max}=9\frac{1000}{32}=281 km$$

My advice - don't let them collide...

• When you say that R = 6400 Km you are rounding correct? because R of earth = 6378.1 Km and then you add the 1000 Km? so R as a total would be 7378.1 Km? Sep 4, 2014 at 0:43
• Yes I was rounding. The radius if the earth depends on where you measure it - it's not a sphere. For the purpose of a question like this, which has lots of unrealistic assumptions, using too many significant figures is counterproductive. And yes $r_p$ would be 7400 km in my example. Sep 4, 2014 at 1:13

I can only assume that the satellites are assumed to meet head-on (rather than at an angle), and that the collision is modeled as inelastic (rather than an explosive collision where pieces fly everywhere).

If so, you have the velocity of each and the mass of each. You can use the conservation of momentum to find the new velocity of the combined-mass object at what will be apogee.

Are you familiar with the vis-viva equation? You can then use that to find other specific points in the orbit. There will be one other point in the orbit that you will want to find.

R will always be from the center of the earth. However you do need to consider the radius of the earth's surface. Do you know why?

So after a bit of research, i came up with this answer. As i had said before we know that in orbit, the mass of the satellites does not matter to find the velocity. so if we use: $$v = \sqrt{GM/R}$$ as M being the mass of earth i can find the final velocity of both satellites after collation. $$Vf = (m2v2 - m1v1)/ (m1 + m2)$$ I came up with 4407.87 m/s

If i take this final velocity back to the first equation, and try to get r, i can compare with the ra and if r is smaller, it means it will crash on earth.

$$r = GM / Vf^2$$

So i find r to be 4527 m according to this, it will definitely will crash to the earth. because the original ra was 7378.1 km

Am i right to assume this?