GREPhysics.NET: GR0177 #22

GR0177 #22

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

Mechanics $\Rightarrow$ }Orbits

The planet orbits in an ellipse, and thus the sum of the minimum and maximum distance is twice the semimajor axis of the orbit. Choice (E) is out.

From Kepler's third law (which can be derived from the formalism of central force theory), one has $T^2\propto r^3$ . Thus, knowing the distances, one can find the period. Choice (D) is out.

The speed can be found from conservation of energy. For an elliptical orbit, one has $E=-k/(2a) = 1/2 mv^2 - k/r$ , where $k=GmM$ and a is the semimajor axis. (One can find the total energy of the orbit from the Virial Theorem. Recall that $\langle T \rangle = -\langle \sum F\cdot r\rangle/2$ . Plug in the gravitational force $F=k/r^2$ to get $\langle T \rangle = \langle k/2r \rangle = \langle k/2a \rangle$ . Thus, $T+V = k/2a - k/a = -k/2a$ .)

The full form of Kepler's third law is $\tau^2 \approx a^3/(GM)$ . Thus, one can determine the mass of the planet.

The only one that remains is choice (A).

Alternate Solutions

Ning Bao
2008-02-01 06:11:15

Consider the asteroid belt. They're not all the same size. Thus, their orbits are not determined by masses, because certainly in the muck there are a few asteroids of wildly varying size with very similar orbits.

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KarstenChu
2007-03-22 14:03:05

It's possible, it seems, to just look at the case of a circular orbit. Setting GMm/r^2=mv^2/r reveals that m cancels out....no mass of the moon for you!

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travis.nicholson
2006-10-23 01:49:09

A more appropriate solution would be:

$2a = r_{min} + r_{max}$ , where $a$ is the semimajor axis. This elimiates choice (E).

The angular momentum of the moon is conserved because there are no external torques. Therefore $l_{min} = l_{max} -> m v_{min} r_{min} = m v_{max} r_{max} -> v_{min} = {v_{max} r_{max}}/r_{min}$ , where $l$ is the magnitude of the angular momentum, and $m$ is the mass of the moon. This eliminates choice (C).

Since there is no external work being done on or by the moon, the mechanical energy of the moon is conserved. Also, since the potential energy of the moon goes as $1/r^2$ , the potential energy is minimal at $r_{max}$ and maximal at $r_{min}$ . Thus, the kinetic energy (and therefore the velocity) is maximal at $r_{max}$ and minimal at $r_{min}$ . With the $v_{min}$ now known (as explained in the previous paragraph), the mass of the planet, $M$ , can be calculated using the conservation of energy:

${m {v_{max}}^2}/2 + {G m M}/{r_{min}}^2 = {m {v_{min}}^2}/2 + {G m M}/{r_{max}}^2$ . Note again that $m$ cancels. This eliminates choice (B).

Now that the mass of the planet is known, the period, $\tau$ , can be calculated using Kepler's third law:

$\tau^2 = {4 \pi^2 a^3}/{G M}$ . This eliminates choice (D).

Only choice (A) remains.

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travis.nicholson
2006-10-23 01:40:58

A more appropriate solution would be:

$2a = r_min + r_max$ , where $a$ is the semimajor axis. This elimiates choice (E).

The angular momentum of the moon is conserved because there are no external torques. Therefore $l_min = l_max -> m v_min r_min = m v_max r_max -> v_min = {v_max r_max}/r_min$ , where $l$ is the magnitude of the angular momentum, and $m$ is the mass of the moon. This eliminates choice (C).

Since there is no external work being done on or by the moon, the mechanical energy of the moon is conserved. Also, since the potential energy of the moon goes as $1/r^2$ , the potential energy is minimal at $r_max$ and maximal at $r_min$ . Thus, the kinetic energy (and therefore the velocity) is maximal at $r_max$ and minimal at $r_min$ . With the $v_min$ now known (as explained in the previous paragraph), the mass of the planet, $M$ , can be calculated using the conservation of energy:

${m {v_max}^2}/2 + {G m M}/{r_min}^2 = {m {v_min}^2}/2 + {G m M}/{r_max}^2$ . Note again that $m$ cancels. This eliminates choice (B).

Now that the mass of the planet is known, the period, $\tau$ , can be calculated using Kepler's third law:

$\tau^2 = {4 \pi^2 a^3}/{G M}$ . This eliminates choice (D).

Only choice (A) remains.

travis.nicholson
2006-10-23 01:50:57

Sorry ... for some reason when I attempted to fix the syntax in this solution the page would not allow me to submit the edited version. Therefore, I posted the solution again.

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Comments

solar39
2009-11-04 09:52:25

When you try solving the period with this equation you're actually solving it for this Lagrangian : rn $\ L=\frac{1}{2}\m_r\dot{r}^2+\frac{GM_pm_m}{r}$ rnwhere $\m_r$ is the reduced mass of the planet and the moon.rnThe Euler-Lagrange equations would give you the equations of motion as: $\m_r\ddot{r}-\frac{GM_pm_m}{r^2}=0$ rnand the energy conservation equations change as:rn $\frac{1}{2}\m_r\dot{r}^2+\frac{GM_pm_m}{r}=E$ rnso i guess you do have the "summation" to solve for the period but you can't get $\ M_p$ and $\ m_m$ seperately.rnI think ETS would have wanted you to just think of this problem as a "Kepler problem in approximate form."

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mlg52
2008-09-27 23:01:00

I take issue with this answer! This is the way I did it.

1. Given Rmin and Rmax you can sum them and divide in two to determine the semi-major axis, a. Thus choice E is eliminated

2. Given Rmin and the semi-major axis you just found, you can compute the eccentricity:
Rmin=a(1-e)
where a is the semi-major axis and e is the eccentricity

3. Given Vperi and Rmin you can calculate the mass of the planet:
Vperi = sqrt{(GM(1+e))/Rmin}
Thus choice B is eliminated

4. Similarly, since Rmax was also measured, you can now find the minimum velocity, which will occur at apastron:
Vap= sqrt{(GM(1-e))/Rmax

Thus choice C is eliminated

Now we are left with the mass of the moon or the period of the orbit. Kepler's law is *actually*

P^2= (4pi^2)a^3/(G(M+m))

Note the denominator contains the *sum* of the masses!

So we truly have one equation left with two unknowns. You *cannot* calculate the *exact* period from here. You can only make an approximation where you neglect the mass of the moon, but that seems like a self-fulfilling prophecy or circular reasoning to me. Saying "a very small moon" is very subjective. What if the planet, too, is very small? They don't state the ratio of the planet to the moon, or mention that it is negligible in size compared to the planet. So how can we really use Kepler's law in approximate form in this case?

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mlg52
2008-09-27 22:58:20

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Ning Bao
2008-02-01 06:11:15

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bootstrap
2007-04-06 19:17:26

There is a very easy approach to this. It involves no math, just a basic understanding of the concept. Lets imagine that its not a "very small moon" and a planet, but a satilite orbiting Earth. It could even be a random rock that got stuck into orbit. Both can have the exact same measurements that are provided by this question. So you cant you tell the difference between a rock and a satilite. Another important think to note is the difference between the mass of the earth and a satilite. Maybe for a heavy object about 10000Kg compared to the Earth or some other planet of the order 10^24Kg. If they didnt use the words "very small" then the mass of the moon is important, and you are able to find it (due to other effect that im not going to get into, b/c i believe you need a computer or a day to do the calculations).

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KarstenChu
2007-03-22 14:03:05

It's possible, it seems, to just look at the case of a circular orbit. Setting GMm/r^2=mv^2/r reveals that m cancels out....no mass of the moon for you!

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merdeonme
2007-03-11 02:35:06

two corrections:

you have:

$L_{min} = L_{max} \Rightarrow mv_{min}r_{min} = mv_{max}r_{max} \Rightarrow v_{min} = \frac{mv_{max}r_{max}}{r_{min}}$

When it should be:

$L_{min} = L_{max} \Rightarrow mv_{min}r_{max} = mv_{max}r_{min} \Rightarrow v_{min} = \frac{v_{max}r_{min}}{r_{max}}$

You also have:

$\frac{1}{2}mv_{max}^2 + \frac{GmM}{r_{min}^2} = \frac{1}{2}mv_{min}^2 + \frac{GmM}{r_{max}^2}$

when it should be:

$\frac{1}{2}mv_{max}^2 + \frac{GmM}{r_{min}} = \frac{1}{2}mv_{min}^2 + \frac{GmM}{r_{max}}$

merdeonme
2007-03-11 02:35:52

Meant this to be a reply to travis.nicholson

jonny23
2007-06-15 02:43:08

It's possible, it seems, to just look at the case of a circular orbit