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Homework Help: Gravity Problems Help

  1. Apr 20, 2004 #1
    I'm stuck on 2 problems.

    T5.11.96) One big question is planetary science is whether each of the rings of Saturn is solid or composed of many smaller satellites. There is a simple observation that can now be made to resolve this issue. Measure the velocity of the inner and outer portion of the ring: if the inner portion of the ring moves more slowly than the outer portion, then the ring is solid; if the opposite is true, then it is composed of many seperate chunks. a) If the thickness of the ring is r, the average distance of the ring from the center of Saturn is R, and the average velocity of the ring is v, show that v(out)-v(in) ~ rv/R if the ring is solid. Here, v(out) is the speed of the outmost portion of the ring, v(in) is the speed of the outermost portion, and v is the average velocity of the ring. b) If, however, the ring is composed of many small chunks, show that v(out)-v(in) ~ -.5(rv)/R. (Assume that r<<R.)

    I tried integration, which didn't work. I tried using linear equations of motion to solve, but that got me nowhere either. This problem is a level 3 question, which is the toughest in the book. I've tried...but to no avail. I know that the rings closer to Saturn have a higher velocity...but that didn't help me much either.

    T5.11.101.) United Federation Spaceship Excelsior is dropping two robot probes to the surface of a neutron star for exploration. The mass of the star is the same as that of the sun, but the star's diameter is only 10 km. The robot probes are linked together by a 1-m-long steel cord, and are dropped vertically(that is, one always above the other). a) Explain why there seems to be a "force" trying to pull the robots apart. b) How close will the robots be to the surface of the star before the cord breaks? Assume that the cord has a breaking tension of 25000N and that the robots each have a mass of 1kg.

    I know this question seems stupid because a neutron star would crush these probes in a matter of seconds. I'd figure since there is so much massin such a small area that the force of gravity would pull down on them more. But I also know that the force of gravity increases with a shorter radius...so wouldn't an object farther out be less affected?
    b) For part b, do I just plug in the equation for gravitational force? I know the answer is 220km...but I'm unsure how to get it. I thought it was more like setting the force to 25000N and then solving for "r^2"...but somehow it went wrong. Am I doing something wrong here or maybe it's just calculations error? Thanks.
  2. jcsd
  3. Apr 21, 2004 #2


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    Actually I thought it was pretty well known that the rings of Saturn are NOT solid! :)

    You don't need any fancy calculus for this. The inner edge of the ring is distance R from the center, the outer edge, distance R+r. A point on the outer ring goes around a total distance of 2&pi;(R+r) in one complete revolution while a point on the inner ring only goes 2&pi;R. If the rings were solid, they would take the same time. That is:
    VOT= 2&pi;(R+r) and VIT= 2&pi;R. Dividing one equation by the other VO/VI= (R+ r)/R= 1+ r/R.

    Number 2 is what is called a "tidal" problem- the question is not just the strength of the gravity but how it changes with distance.

    F= -GMm/r2 but there are two different "r"s! Assume the nearer of the two masses is distance r meters from the center of the star. The other would be distance r+ 1 meters. The force on the nearer would be -GMm/r2 and the on the farther -GMm/(r+1)2. What is the difference between the two forces? When will that difference be 25000 N. (You will need to get "-GM" from the fact that the mass of this star is the same as the sun.)
  4. Apr 24, 2004 #3
    I got the first part of problem 1, but the 2nd part of problem 1 is bugging me. How can you actually know what the velocity(in symbolic form) is?

    You said for part b of the 2nd problem was like, set 25000 = -GmM/r^2...but I got some gigantic number(about 73000km) instead of 220. I swapped both F and r^2, but no cigar. Thanks for being so helpful though. These 2 problems have been bugging me all week.
  5. Apr 25, 2004 #4


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    I didn't say the forces would be 25000, I said the difference in forces would be 25000!

    (By the way, what are you getting for GM?)
  6. Apr 25, 2004 #5
    I got G = 6.67E-11, and M=2E30 kg and just multiplied GM together.

    But I think I might get it now...it's the first problem of part b that's bugging me. Thanks. :)
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