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?
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).
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?
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.
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?
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.
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
Hi Esther. Before the collision you have one mass with angular velocity ω and another mass with angular velocity -ω. So (if the radius of orbit is r) the total momentum is mrω – mrω (which is zero).
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).
I really appreciate your clear answers for the rest of the question, though. They’re very helpful.
Thanks very much indeed for considering!
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))
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.
Fill in your details below or click an icon to log in:
You are commenting using your WordPress.com account. ( Log Out / Change )
You are commenting using your Twitter account. ( Log Out / Change )
You are commenting using your Facebook account. ( Log Out / Change )
You are commenting using your Google+ account. ( Log Out / Change )
Connecting to %s
Notify me of new comments via email.
Notify me of new posts via email.