Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Orbital Mechanics Quandary

  1. Mar 11, 2008 #1
    I had thought that for an object in a simple circular orbit, the escape velocity would be root 2 times it's orbital speed. And that this should approximately apply for the Moon too. However Wikipedia here


    gives the moon's escape velocity as 2.38 km/sec, yet further down gives the "average orbital speed" as 1.022 km/sec. So, is the Root 2 relationship valid?

  2. jcsd
  3. Mar 11, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    Orbital speed (of the moon) is not the same parameter as escape velocity. Orbital speed is the linear (tangential) speed of a satellite, moon or planet around the body that it orbits. Escape velocity is the speed required for a mass to depart from some point in the gravitational field, e.g. the surface of moon or planet, or orbit, and escape to some very far distance (essentially infinitely far away), such that if the propulsive force is cutoff, the mass would coast to infinity.
  4. Mar 11, 2008 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I think you'll find that the 2.38km/sec value given is the escape velocity for an mass resting on the Moon's surface, not the escape velocity needed for the Moon to leave Earth orbit.
  5. Mar 13, 2008 #4
    thanks Janus, that explains it. I used Vesc = sq root(2*G*M of moon/R) and calculated the escape vel for an object on the surface of the moon to be 2.375 x 10E3 m/sec, almost exactly what wikipedia states. Further, I used the same equation to calculate the escape velocity of the moon from the Earth (assuming no other gravity factors), and came up with 1.440 x 10E3 m/sec, which works out to be almost exactly sq root of 2 * the "average orbital speed" given by wikipedia. So the sq root 2 relationship between orbital speed, and escape speed seems correct.

    Now my real ambition is to verify that the moon cannot escape (all other gravity factors aside), that there is not enough kinetic energy in the Earth's rotation. Assuming that it were to be all transfered to the moon (I know tidally locked still means the Earth has a small rotation but lets not consider this for now) I'd like to prove that the moon would still be in orbit. I realize that the sun will have probably flared up and died before this could theoretically happen, but it's just an academic exercise.

    For the orbital mechanics gurus out there, would it be correct to calculate the Earth's rotational energy, add that to the the moon's potential energy, and then see if that's greater than the energy needed for the moon to escape?

    Or this alternative which I believe gives the same answer. We know the moon's escape vel from the Earth is sq root of 2 * orbital vel, and thus the escape energy needed is 2 * orbital kinetic energy. And so if the Earth's rotational energy is less than the moon's kinetic energy, then that must mean the moon can not escape?

    Is this too simple, or are there any other factors to consider that would affect this calculation?

    Should I go ahead with my assumptions?
  6. Mar 13, 2008 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    A couple of things.

    Actually, you need to calculate the total energy of the Moon (kinetic and gravitational), then compare that to the rotational energy of the Earth. If the absolute value of the Earth's energy is less than the Moon's total energy, then it doesn't have enough energy to boost the Moon to escape velocity.

    The second thing is that the Moon doesn't have to reach escape velocity to leave earth orbit. Because the influence of the Sun's gravity, there will be a point where the Sun's grip on the Moon will be strong enough to pull the Moon free into a independent orbit. This point falls on what is called the Hill sphere. If the Moon drifts further from the Earth than the radius of the Hill sphere, it will drift away with out every reaching escape velocity. So you really need to figure out how much energy it would take to lift the Moon to an orbit with an radius outside the Hill sphere. This will be less than that needed to reach escape velocity.

    A couple of formulas to help you along

    total energy of a orbiting body:

    [tex]E_t = -\frac{GMm}{2a}[/tex]

    M = mass of primary
    m = mass of orbiting body
    a= average orbital radius of orbit. (semi-major axis)

    Hill sphere radius for circular or nearly circular orbits:

    [tex]r = a \sqrt[3]{\frac{m}{3M}}[/tex]

    m = mass of secondary body
    M = mass of primary body

    for example to find the Hill sphere for the Earth, The Earth is the secondary body and the Sun the primary. The answer will the be the maximum distance from the Earth at which a body can stay in orbit around the Earth.
  7. Mar 15, 2008 #6
    So Janus, using your formula for the moon’s total energy (other than some small rotational energy about it’s center), I get:

    E = -6.6726 E-11 * 5.974 E24 * 7.348 E22 /(2*3.844 E8) = -3.81 E28 Joules

    Now for the rotation energy of the Earth, I used Ek = mv^2/5
    Where m is the mass of a spinning solid sphere, and v is the equatorial velocity.

    I calculate Ek = 2.57 E29 Joules, whereas wikipedia lists it as 2.6 E29 Joules, very close.

    So the moon would need 3.81 E28 joules to escape (neglecting other gravity influences for now), and Earth has 2.57 E29 joules of rotational energy. So the Earth has approx 7 times as much rotational energy as needed by the moon to escape.

    Now all sources I’ve seen say that theoretically there’s not enough energy in the Earth for the moon to escape, that tidally locked would be the theoretical answer, and never escape the Earth (caveats listed in my post #4).

    So where’s my error?
  8. Mar 16, 2008 #7


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The problem is in assuming that all of the Earth's rotational energy can be transfered to the Moon. Only a small percentage is.

    Consider the following:

    The Earth rotation slows by 1 millisecond per century.
    The moon recedes at a rate of 4 cm per year.

    If we figure out how much energy the Earth loses in rotational energy per century and then figure out how much energy the moon gains by receding 4 meters during that century we can determine what fraction of the energy the Earth loses is transfered to the Moon. The answer will come out to be smaller than 1/7. The rest is lost as waste heat.
  9. Mar 17, 2008 #8
    Here’s what I calculated:
    The Earth’s loss in rotational energy due to a change of one millisecond (happens over a century) is 6.00 E21 joules
    The moon’s gain in energy receding 4 m, (receding 4 cm per year ) is 3.96 E20 joules

    So the fraction of the energy the Earth loses compared to the moon gain is
    6.00 E21 / 3.96 E20 = 15

    As you predicted, the moon gains less than the one seventh required to escape. One fifteen seems a pretty small amount of energy to transfer. I guess “global warming” gets the other 14/15ths.

    So... using the 1/15th factor, how far would the moon recede, outside gravity and longevity considerations aside?
  10. Mar 18, 2008 #9


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Add 1/15 of 2.57e29 joules to -3.81e28 joules,giving you the new total energy of the Moon. Take this answer and plug it into the equation used to find the total energy of the Moon and solve for "a" the new average radius of the orbit.
  11. Mar 22, 2008 #10
    So the ratio of the energy the Earth loses compared to the moon gains is
    6.00 E21 / 3.96 E20 = 15.14

    Total energy of the moon after receiving 1/15th of the Earth’s rotational energy = -2.11 E28 joules.

    Radius for that energy = 694,000 km, compared to 384,400 present radius of moon’s orbit.

    Moons period at a radius of 694,000 km would be 66.6 present Earth days (compared to present 27.4 or so).

    Hill sphere radius for the Earth/Sun works out to be 1,497,000 km. So the sun should never capture the moon.

    For an orbiting object at the Hill sphere radius I calculated, the sun’s force of gravity on this object would be 33 times the Earth’s force of gravity on it. Yet at just less than this radius, the object would theoretically stay in an (un)stable orbit around the Earth. This was calculated from

    Fg = G*M1*M2/Rsquared.

    Does this raise any eyebrows?
  12. Mar 22, 2008 #11

    Ken G

    User Avatar
    Gold Member

    That all seems right, including Janus' point about the energy lost to wasted heat. The one issue that has not been explicitly mentioned however is that the actual conserved quantity, to a good approximation, here is angular momentum not rotational energy. The Earth loses its angular momentum of rotation as the Moon gains orbital angular momentum. As the Moon would require infinite angular momentum to escape the Earth, it would never be possible for any amount of energy. However, the Sun does ruin the angular momentum conservation at some point, controlled by the Hill sphere as mentioned above. So putting it all together, one might say that the Earth has plenty of rotational energy to eject the Moon, but it lacks sufficient angular momentum to get the Moon far enough away where the Hill sphere effects become important. The Earth's rotation would slow to the point that it would be locked to the Moon's orbit, and at that point there would no longer be any reason for the Earth to slow down further. In fact, IIRC, the effects of the Sun are then to cause the Moon to start coming back in toward the Earth.
    Last edited: Mar 22, 2008
  13. Mar 22, 2008 #12


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Try calculating the ratio of the Sun's force of gravity to the Earth's on the Moon in its present orbit.
  14. Mar 22, 2008 #13
    Janus, I did previously calculate the ratio of the sun's force of gravity on the moon compared to the Earth's, in its present orbit, and came up with 2.21

    Should there be a theoretical maximum for the ratio of forces of gravity at the Hill sphere for different primary and secondary objects?
  15. Mar 22, 2008 #14


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The point is that is that is not the ratio of the forces of gravity that counts. Instead you need to consider the ratio of Earth's gravity to the tidal force of the Sun acting across the orbit of the Moon.
  16. Mar 26, 2008 #15
    Ken, I understand that in the end angular momentum is conserved (somehow), however I’m finding my calculations are NOT showing this. For the situation where the primary (the Earth) slows 1 msec over 100 years, I’m finding the change in the rotating Earth’s angular momentum L = 2/5 * Me * Re * (Vi – Vf) = 8.23 E25 Nms.

    I considered two cases, the first where ALL of the Earth’s kinetic energy loss (due to decrease rotational speed) goes to the moon (an academic example only). The moon’s change in L = Mm * (G * Me)^.5 * (Rf ^.5 – Ri ^.5) = 2.27 E27 Nms, where Rf is the lunar distance after absorbing the Earth’s Ek loss, and Ri is the lunar distance at the start (the difference being about 60 m).

    For the second case (using Janus’ realistic numbers of .001 msec over 100 years, and 4 cm per year) I calculate the moon’s change in L to be 1.47 E26 Nms (almost exactly 1/15 th of the first case angular momentum, as expected).

    However, neither of these cases equals the smaller calculated Earth L loss of 8.23 e25 Nms. Conservation of energy was explained by creation of heat energy. Is there a similar explanation for conservation of angular momentum ?

    And Ken, would an escaping moon (Hill sphere and other factors aside) really require infinite angular momentum?
  17. Apr 7, 2008 #16
    Sorry if this simple way of putting the answer to spud three's initial problem has been mentioned already.

    The two parameters that are root 2 apart can be found without any mention of the Earth. Escape velocity at the surface of the moon is root 2 larger than orbital velocity at the surface of the moon. You'd hit a few hills flying that low, but the point is that wikipedia gave the escape v at the surface, so r is the moon's radius, to the surface, in both

    root GM/r, and root 2GM/r. Hope this way of seeing it makes sense.
  18. Apr 14, 2008 #17
    ...Back to Ken's angular momentum...

    Yes thanks Jenk, I came up with the similar formulas and derivation to show root 2 once Janus explained what Wiki meant, and so the quandary referred to in the post title is solved for me.
    However, the thread has progressed to a disussion of conservation of angular momentum, as in Ken's points in post 11 and my attempts in post number 15. I've seem to find that in the 100% energy transfer case (assumed in my post number 6) angular momentum seems not to have been conserved. Similarly for the partial energy transfer mentioned by Janus in post number 7 and calculated by me to be approximately 1/15 th in post 8, angular momentum again seems not to be conserved.
    This I worked out for the Earth using I = .4 * M * R squared, and angular momentum L = I * w. And for the moon using L = M * V * R. (I can post the complete calculation if anyone would like to take a look and offer me an explanation of where the missing angular momentum is).
    In further pursuit, can ANYONE offer some figures on a ficticious planet/satellite system that undergoes changes in the satellite orbit (for example as tides cause on the Earth/moon system) where I can then calculate the angular momentums?

  19. Apr 14, 2008 #18

    Ken G

    User Avatar
    Gold Member

    Sorry I missed this before. Conserving energy won't conserve angular momentum, that's no suprise (the same happens with energy and linear momentum when two cars collide). The second example probably just used approximate numbers, and that's why the angular momentum wasn't conserved exactly. Note it wasn't too far off.
    It wouldn't really require an infinite angular momentum if you include the Hill sphere, only in the idealized limit of two point masses and no Sun. But it would require more angular momentum than is present in the Earth's rotation, so that's the key issue. As for why angular momentum is conserved, it is because the Earth-moon system (with no Sun and no Hill sphere) is a closed system with no external torque on it.
    Last edited: Apr 14, 2008
  20. Apr 18, 2008 #19
    Well Ken, using the figures over 100 years of the moon receding 4 m and the Earth slowing 1 milli second may be just too approximate, as you say. Because using them, it appears the Earth would have to lose 1.823 times as much angular momentum as I calculated, in order to conserve angular momentum. Hence if we revise the Earth's rotation slowing to 1.823 milliseconds over the 100 years, then the moon/earth angular momentum is conserved by my calculations.
    Then as for conservation of energy, since the Earth is slowing almost twice as quickly as the previously used figure, instead of approximately one fifteenth of the Earth's lost kinetic energy going to the moon, it's more like 1/(15.14*1.823) = .00363 or approx 1/28. The remaining 27/28 must go to heat then.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook