# Oxford PAT 2012, Question 21

## 14 thoughts on “Oxford PAT 2012, Question 21”

1. rob says:

You said that KE is proportional to 1/r, by looking at the integral of Force w.r.t. distance. But if you evaluate this integral between 2 bounds, say a and b with a < b, then KE=2GMm[(b-a)/ab], which is not an inversely proportional relationship.
Doesn’t the inverse proportionality only occur when the object has fallen from (something approximating to) infinity to the desired distance r from the centre?

1. Hi Rob. Thanks very much for that comment — you are absolutely right in your reasoning, and you make a very good point. As you say, if the point of collision of the meteoroids were at some very large distance (near infinite) from Earth, then the subsequent KE would be proportional to 1/R. But in the question it says that the collision is at a distance of ‘a few … times’ the radius of the Earth; infinity is normally considered to be a bit bigger than ‘a few’!

But now we are left with the question of how we go about sketching the graph. I think that 1/R is still a reasonable approximation to the shape of the graph, so I would be inclined to sketch the graph the same way, but in my answer to make it clear that the 1/R is just an approximation, as you pointed out. What do you think? Maybe there is another expression we could use that would be more accurate, but still simple enough for a ‘sketch’.

2. Max Leftley says:

Can I just check I understand your answer to the first bit involving momentum? We haven’t done angular momentum in Maths yet so I don’t know if this is right.. Would it be possible for you to direct me on how best to go about looking at it?
So anyway… my line of thought about what you put is that when the two meteoroids collide, as the initial angular momentum was zero, the final angular momentum must be too – and the azimuthal angle therefore doesn’t change as the combined mass falls with no external forces acting on them? Because there is no ‘orbital’ velocity/momentum anymore?

1. Yes that’s basically right. The two masses have equal and opposite momentum, so the total momentum (before and) after collision is zero. After the collision the meteoroids could do two things: either bounce off and move away in opposite directions at the same speed, or coalesce. The question tells us that in this case they coalesce, so the coalesced mass must end up being stationary (until it falls to earth).

1. Max Leftley says:

Thank you!

3. Alicia says:

Hi, I was wondering why the temp increased after 2Re, since u mentioned cooling. I thought temperature in space is very much lower than that of the atmosphere?

1. Yes that’s right, but the mass is falling to Earth and, once the mass hits the atmosphere at distance 2Re, its temperature rises because air resistance slows it down and the energy dissipated heats up the mass. (See the answer above) this means that below Re, as R increases temperature decreases.

4. Alicia says:

Hi, I was wondering why the temp increases as R increases? Since u mentioned cooling, I expected the graph to curve downwards. Furthermore, I thought temp in space is very much lower than the temp near earth?

1. Hi Alicia, thanks for the question. You make a really interesting point; the source of the confusion is that the graph is of temperature against distance, and so we don’t see the passage of time on the graph.

What is actually happening is that a hot mass is falling towards the Earth. As it falls, the mass cools down, so as R decreases, T decreases; but this means that as R _in_creases, T _in_creases.

Yes the temperature in space is very much lower than on the Earth, so the temperature could fall to a very low level, but then the mass heats up very quickly when it hits the atmosphere.

5. Esther says:

Hi, I don’t completely understand the second part of the question-how is momentum conserved in this was with zero angular velocity? Thank you so much for the advice and help on this website by the way, it has been an invaluable resource for me

1. Hi Esther. Before the collision you have one mass with angular velocity ω and another mass with angular velocity . So (if the radius of orbit is r) the total momentum is mrω – mrω (which is zero).

6. ghghuy says:

Hello, I have a slight issue with the first part of the question – I was always taught that orbital speed does not depend on mass of the satellite. I think you might have just written the equation for gravitational force incorrectly by making it (M+m) instead of (Mm).
Thanks very much indeed for considering!

1. Peter says:

Thats what I figured to: Using the equation for the centripetal force Fz = m * v^2 * r^(-1) and the gravitational force Fg = G * M * m * r^(-2) I get:

Fz = Fg
=> m * v^2 * r^(-1) = G * M * m * r^(-2)
<=> v^2 = G * M * r^(-1)
<=> v = squareRoot(G * M * r^(-1))

2. Yes you are basically right that in practice the gravitational acceleration (and hence orbital speed) is independent of the mass of the satellite, but this is only an approximation. In reality, the acceleration is proportional to M+m, and not M (i.e. the force is (M+m)m, and not Mm), but because the mass of the earth is much bigger than the mass of the satellite, you can normally ignore the m in the term M+m.

Have a look at this: https://oxfordpat.wordpress.com/circular-orbits-under-gravity/, which goes into a bit more detail.