Question on Orbits and Kepler/Uni. Grav. Laws

  • #1
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Homework Statement


The picture has the problem question.
image1.JPG


OK.

When the moon is at point B, the distance from the moon to the center of the planet is most nearly:

A) (1/25) * rA
B) (1/5) * rA
C) (1/ sqrt(5) ) * rA
D) rA
E) sqrt(5) * rA.

Homework Equations



newton's law of gravitation. kepler's third law

The Attempt at a Solution


[/B]
Ok. I am told that the correct answer is B but I do not know why.

I got A instead. Here's what I did.

I solved from my rA for the planet at point A.

I apologize for my bad math formatting, but I am in a rush and can not use latex. Hope you understand.

Gravitational force causes centripetal acceleration:

G * mPlanet * mMoon mMoon * (v0)^2
--------------------------- = -----------------------
(rA)^2 rA.

I cancel mMoon and one of my rA terms and I solve for v0^2. I got v0^2 to be equal to
G * mPlanet/r.

I then did the same thing for my planet in position B.
G * mPlanet * mMoon mMoon * (5v0)^2 mMoon * 25 * v0^2
--------------------------- = ----------------------- = ----------------------
(rB)^2 rB. rB

I plugged in my calculated result for v0^2 for position A into the expression and solved for rB.
My G and my mPlanet terms cancel. I got rB = rA/25.

Where did I mess up?

Thanks in advance.
 

Answers and Replies

  • #2
Hi,

I'm absolutely not an expert in that field but I may have another approach:

Maybe you should consider the second Kepler's law: the conservation of the area in a given time.

The moon travels v0*t in the time t, if you consider a small enough time, you have the small angle approximation which provides tan(theta)=theta so the area is vo*t*rA

This value has to be equals to 5*v0*t*rB, so rA is 5 times bigger than rB....

Is it ok for you ?

Regards,

Anton
 
  • #3
410
11
I read up a little on kepler's second law but they did not teach it in my high school class, so I am afraid to use that approach, especially with the small angle approximation.

I do see how your approach works and makes sense... but my AP Physics exam is looking for the third law, and I am trying to use it, if possible.

Thanks a lot for that interesting thought!
 
  • #4
gneill
Mentor
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The problem is that for elliptical orbits the centripetal force does not balance the gravitational force. That's why the planet's speed changes and the orbit is not circular.

A better approach is to use a conservation law. Kepler's 2nd law is really a statement about conservation of angular momentum, and you should look at the problem from that point of view. The two points of interest in this problem (apogee and perigee) are particularly amenable to this approach (why?).
 
  • #5
410
11
Ahh. I see the pitfall now.

Kepler's 2nd law makes sense now.

I see why kepler's third does not make sense.

Thanks!
 
  • #6
410
11
but is there any way I can use Kepler's third at all? modified, perhaps?
 
  • #7
I plugged in my calculated result for v0^2 for position A into the expression and solved for rB.
My G and my mPlanet terms cancel. I got rB = rA/25.
Between 25 and 5 you clearly have squared something by mistake, I haven't done the calculations but you should check if you didn't consider v0² instead of v0 somewhere...
 
  • #8
gneill
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but is there any way I can use Kepler's third at all? modified, perhaps?
If there is it would probably be a very round-about way. You have to work with the information given, which does not directly give you the length of the semimajor axis of the orbit or the period. Nor are you given any actual mass values or velocities, and you aren't even supplied with an eccentricity value to help find other related parameters, so the Newton form of Kepler III is no help.

I think this problem is constructed specifically to be solved using conservation of angular momentum.
 
  • #9
SammyS
Staff Emeritus
Science Advisor
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but is there any way I can use Kepler's third at all? modified, perhaps?
Please state Kepler's Third Law for us.
 
  • #10
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11
Ok. I see how kepler's third law would not work. and I took (5 times v0)^2 and wrote it as 25 times v0^2. i do not think i made a mistake there.

kepler's third law is that the period squared is proportional to radius of orbit cubed.
 
  • #11
410
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and they did state the masses to be m and 49m, so I could use them, no?
 
  • #12
gneill
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and they did state the masses to be m and 49m, so I could use them, no?
Well, yes and no. A simple ratio of masses for primary and satellite won't tell you much about the orbit since that depends on the total mechanical energy and the eccentricity. Kepler III gives you a handle on comparing the periods and sizes of different orbits in a system, but not much about one single orbit. You might be able to do something with the complete Newton version of Kepler III and define your own mass unit and a gravitational parameter μ (equivalent to GM) to fit the system. Then you'd still need to use an equation other than Kepler to relate velocity and radius for a single orbit.
 
  • #13
410
11
hmmm... that sounds a bit over my head with my limited knowledge of physics. thanks a lot, though!
 

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