Oxford PAT 2012, Question 21

2012_page19

 2012_page20

9 thoughts on “Oxford PAT 2012, Question 21

  1. 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.

  2. 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.

  3. 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).

  4. 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!

    1. 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.

Add your comments

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s