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If the moon were shrunk to having a radius of one meter, without changing its distance (of the apogee) from the earth, what would be the slowest it could be slowed down to and still orbit the earth, albeit in a highly elliptical orbit? It shouldn't hit the earth. The speed at the apogee is what is sought. The apogee must be at 400,000 km from the earth, same as the moon's center is now.

The moon orbits the earth at close to one kilometer per second in a fairly round orbit. So the answer must be considerably less than that speed.

Someone better at math than me might be able to calculate it from the equations on this page: https://en.wikipedia.org/wiki/Vis-viva_equation.

The lunar distance is approximately 400,000 km says https://en.wikipedia.org/wiki/Lunar_distance_(astronomy)

The radius of the earth is approximately 6400 km says Wikipedia: "A globally-average value is usually considered to be 6,371 kilometre" https://en.wikipedia.org/wiki/Earth_radius.

The nearest to the surface of the earth that a one meter radius lump of moon rock could fly and stay in orbit, I estimate to be 100 km.

Matthew Christopher Bartsh
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  • "This question doesn’t meet a Physics Stack Exchange guideline." What guideline? – Matthew Christopher Bartsh May 01 '21 at 04:07
  • Why do you say that the size of the moon is relevant to its orbit? – Sandejo May 01 '21 at 04:17
  • This question, in its current state, is a homework-like question. Please see https://physics.meta.stackexchange.com/q/714 – PM 2Ring May 01 '21 at 04:19
  • @Sandejo The bigger the moon, the faster it needs to be going at the apogee to avoid colliding with the earth at perigee. It has to squeeze by the earth, as it were. – Matthew Christopher Bartsh May 01 '21 at 04:20
  • By bigger, do you mean bigger in size or bigger in mass ? Like someone already pointed out, the mass of a satellite is irrelevant for its velocity/altitude – silverrahul May 01 '21 at 06:05
  • " The nearest to the surface of the earth that a one meter radius lump of moon rock could fly and stay in orbit, I estimate to be 100 km. " What factor are you considering limits it to this ? What would stop an orbit with nearest distance of 70 km ? Is it friction ? – silverrahul May 01 '21 at 06:08
  • @PM2Ring I had a look at that page you linked to. This part was particularly instructive: "Don't forget: closing a question is mostly a subjective decision. The reviewers see a "close" and a "leave open" button next to your question. Although there are site rules, it is mainly up to them how to interpret these rules." Do you have any specific suggestions as to how I could edit the question to prevent it being closed? – Matthew Christopher Bartsh May 01 '21 at 06:29
  • @silverrahul Bigger in radius; Yes, friction. And not being a round number(just my preference). – Matthew Christopher Bartsh May 01 '21 at 06:39
  • Well, questions asking for a calculation (or to check if a calculation is correct) tend to get closed. So you need to ask a conceptual question that will help you solve this problem. However, as you mentioned, you just need to use the vis-viva equation to calculate the apogee speed. So I'm not sure how you can edit this question to improve its chances of staying open. FWIW, the calculator built into Google is great for doing calculations like this. OTOH, so far you only have 2 close votes... – PM 2Ring May 01 '21 at 07:20
  • BTW, I think you need a higher altitude. The ISS orbits at an altitude of ~400 km, and it has to burn fuel on a regular basis to compensate for atmospheric drag. See https://en.wikipedia.org/wiki/Low_Earth_orbit#Orbital_characteristics & https://www.nasa.gov/mission_pages/station/expeditions/expedition26/iss_altitude.html – PM 2Ring May 01 '21 at 07:25
  • FWIW, here's how to do the apogee speed calculation in Google, using a perigee altitude of 400 km = 6778 km perigee radius and $\mu_\oplus$ from https://en.wikipedia.org/wiki/Standard_gravitational_parameter sqrt((3.9860044E14 m^3/s^2)*(2/(406700 km)-1/(206739 km))) – PM 2Ring May 01 '21 at 07:30
  • I pasted it in to the Google search box and Google returned sqrt(((3.9860044E14 (m^3)) / (s^2)) * ((2 / (406 700 km)) - (1 / (206 739 km)))) = 179.255041 m / s which is impressive. The answer is believable although I had guessed about 50 m/s. Now all I need to do is figure out what the question means. – Matthew Christopher Bartsh May 01 '21 at 21:44

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The size of a satellite does not matter for the period/velocity of its orbit. In fact, as long as the mass is much lower than the planet's, the satellite's mass doesn't matter either. So the moon would orbit at about the same speed if it was shrunk to 1 meter.

qazwsx
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By making the Moon’s orbit highly elliptical - in other words, by making the semi-major axis $a$ sufficiently large - you can make the Moon’s velocity at apogee as small as you like, and its orbital period as large as you like. Since you can stretch the orbit in this way while keeping the perigee (the closest point to the Earth) the same as the current Earth-Moon distance, the size of the Moon is irrelevant.

(This, of course, treats the Earth-Moon as a two body system and ignores the effect of the Sun and other bodies in the Solar System)

Note: this answer was written in response to the original question, not the edited version which adds the constraint that apogee is unchanged.

gandalf61
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Okay, so the answer is a fairly straight forward application of the Vis-viva equation.

So, since you want perigee to be (100+6400) km and apogee to be 400000, hence the semi major axis becomes a = 1/2 * ( 6500 + 400000)= 203250 km

You want the velocity at apogee. so the distance to satellite, r = 400000 km

Put these values of a and r in the equation,

and you get v = 178.51 m/sec

Hence, you would need to slow moon down to about 178 m/sec at its apogee to have its orbit pass at a distance of 100 km to earth's surface .

This is of course under the assumptions of that equation which assumes there is negligible air friction at 100 km, but that may not be a valid assumption in this case.

silverrahul
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  • FWIW, using 406700 km for the apogee & 6378+400=6778 km for the perigee, I get 179.255 m/s for apogee speed & 10.756 km/s for perigee speed. And 10.8276 days for the period. – PM 2Ring May 01 '21 at 07:40
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Is what you're asking this? "What would be the orbital velocity at apogee of a point mass that has the moon's apogee but just barely misses the Earth at perigee?"

Then you can use the vis viva equation from the page you linked. The semi-major axis $a$ would be half of the sum of the lunar apogee and the Earth's radius (including your 100 km). Put in the apogee for $r$ to get your answer.

The equation is for "small" mass satellites so it would probably not be entirely accurate. I'm not sure where to go for a better approximation.

Kristoffer Sjöö
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    That formula is fine for a satellite with a tiny mass like this. It's not quite correct for the Moon because the Moon's mass is relatively large. If you assume the Moon has negligible mass, you get a mean orbital speed that's roughly 0.5% too small. – PM 2Ring May 01 '21 at 07:48
  • @PM2Ring That is very interesting. I did not know that. What would be the formula to use when the Moon has appreciable mass? – Matthew Christopher Bartsh May 03 '21 at 19:50
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    @MatthewChristopherBartsh There's a standard way to reduce the two body problem in Newtonian gravity, see https://en.wikipedia.org/wiki/Two-body_problem Also see https://physics.stackexchange.com/q/3534/123208 – PM 2Ring May 03 '21 at 20:04
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    @MatthewChristopherBartsh FWIW, the Moon's orbit is quite difficult to model accurately because its mass is so large relative to its primary (i.e., the Earth), compared to most other satellites in the Solar System. See https://en.wikipedia.org/wiki/Lunar_theory – PM 2Ring May 03 '21 at 20:08
  • @PM2Ring Amazing that binary stars of equal mass 'typically' move like that, both with elliptical orbits, alternately near to each other and moving fast, and far apart moving slowly. I did not know that. Wikipedia has a lot of great animations. – Matthew Christopher Bartsh May 03 '21 at 20:56