Astronomy question
Astronomy question
The moon is edging away from the earth. Does this mean that its velocity is slightly more than escape velocity?
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Re: Astronomy question
They are related, but basically no.
The escape velocity is √2 times the orbital speed, so is always about 41% larger.
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Re: Astronomy question
No. It's velocity is the orbital velocity required to orbit the earth every 27.32 days.
It's orbit is getting slightly longer in period as the Earth's rotation is slowed by tidal drag, and the angular momentum of that is transferred to the Moon's angular momentum about the Earth. But there's not enough momentum/energy there to get the Moon out of Earth's orbit even once they become tidally locked.
It's orbit is getting slightly longer in period as the Earth's rotation is slowed by tidal drag, and the angular momentum of that is transferred to the Moon's angular momentum about the Earth. But there's not enough momentum/energy there to get the Moon out of Earth's orbit even once they become tidally locked.
Re: Astronomy question
No, that's not correct, because orbits at different altitudes have different speeds.
Kepler's third law says that the cube of the period of the orbit is proportional to the square of the semi major axis a, which for circular orbits is the length of the orbit divided by 2 pi. And so the orbital speed is proportional to 2 pi a/a⅔ ~ a⅓
Last edited by dyqik on Fri Jan 21, 2022 12:26 am, edited 1 time in total.
Re: Astronomy question
Fat fingered the quote button instead of the edit button.
Re: Astronomy question
Yes, but the escape velocity from a particular orbit also depends on the altitude.dyqik wrote: ↑Fri Jan 21, 2022 12:25 amNo, that's not correct, because orbits at different altitudes have different speeds.
Kepler's third law says that the cube of the period of the orbit is proportional to the square of the semi major axis, which for circular orbits is the length of the orbit divided by 2 pi. And so the orbital speed is proportional to 2 pi a/a⅔ ~ a⅓
My avatar was a scientific result that was later found to be 'mistaken' - I rarely claim to be 100% correct
ETA 5/8/20: I've been advised that the result was correct, it was the initial interpretation that needed to be withdrawn
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ETA 5/8/20: I've been advised that the result was correct, it was the initial interpretation that needed to be withdrawn
Meta? I'd say so!
Re: Astronomy question
Escape velocity is the velocity you need to reach to escape from a body to infinite distance. It's fixed by the mass of the body. You are talking about the delta vee required to reach escape velocity from a given orbit.Gfamily wrote: ↑Fri Jan 21, 2022 12:29 amYes, but the escape velocity from a particular orbit also depends on the altitude.dyqik wrote: ↑Fri Jan 21, 2022 12:25 amNo, that's not correct, because orbits at different altitudes have different speeds.
Kepler's third law says that the cube of the period of the orbit is proportional to the square of the semi major axis, which for circular orbits is the length of the orbit divided by 2 pi. And so the orbital speed is proportional to 2 pi a/a⅔ ~ a⅓
Re: Astronomy question
Well, I assumed that Grumble's question related to the Moon and the velocity to escape from where it is in orbit, rather than the escape velocity if leaving from the Earth's surface.dyqik wrote: ↑Fri Jan 21, 2022 1:08 amEscape velocity is the velocity you need to reach to escape from a body to infinite distance. It's fixed by the mass of the body. You are talking about the delta vee required to reach escape velocity from a given orbit.Gfamily wrote: ↑Fri Jan 21, 2022 12:29 amYes, but the escape velocity from a particular orbit also depends on the altitude.dyqik wrote: ↑Fri Jan 21, 2022 12:25 am
No, that's not correct, because orbits at different altitudes have different speeds.
Kepler's third law says that the cube of the period of the orbit is proportional to the square of the semi major axis, which for circular orbits is the length of the orbit divided by 2 pi. And so the orbital speed is proportional to 2 pi a/a⅔ ~ a⅓
My avatar was a scientific result that was later found to be 'mistaken' - I rarely claim to be 100% correct
ETA 5/8/20: I've been advised that the result was correct, it was the initial interpretation that needed to be withdrawn
Meta? I'd say so!
ETA 5/8/20: I've been advised that the result was correct, it was the initial interpretation that needed to be withdrawn
Meta? I'd say so!
Re: Astronomy question
Ignore
Re: Astronomy question
My maths is way off here. Orbital velocity is proportional to 1/√adyqik wrote: ↑Fri Jan 21, 2022 12:25 amNo, that's not correct, because orbits at different altitudes have different speeds.
Kepler's third law says that the cube of the period of the orbit is proportional to the square of the semi major axis a, which for circular orbits is the length of the orbit divided by 2 pi. And so the orbital speed is proportional to 2 pi a/a⅔ ~ a⅓
Gfamily, sorry I also seem to be way off of wikipedia's definition of escape velocity which agrees with your usage. Which is weird, because I've never seen escape velocity defined relative to an orbit, despite a doing a PhD in astrophysics.
Re: Astronomy question
The escape velocity corresponds to the energy required to get you out of the earth's energy well. If you're gone from the earth's surface to orbit, you've already use a fair bit of energy to get there, so your remaining energy requirement is lower, so your escape velocity is lower.
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Re: Astronomy question
Your kinetic energy is completely different between being "stationary" on the surface on the earth, and if you were an orbit around the earth's mass at the radius of the earth, so the escape velocity calculation is different based on which assumption you start with.Sciolus wrote: ↑Fri Jan 21, 2022 9:45 amThe escape velocity corresponds to the energy required to get you out of the earth's energy well. If you're gone from the earth's surface to orbit, you've already use a fair bit of energy to get there, so your remaining energy requirement is lower, so your escape velocity is lower.
Basically the kinetic energy you need to have in order to be in an orbit at a given radius is already half the energy you need to get out of the gravity well to infinity. Hence you need to multiply your orbital velocity by sqrt(2) in order to double your kinetic energy.
(Assuming circular orbits for simplicity.)
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Re: Astronomy question
I can't remember where I first came across the two speeds expressed as a direct proportionality (I have a feeling it would have been an Isaac Asimov essay read 45 years ago), and until I looked it up yesterday I'd forgotten that it was a √2 constant term.
My avatar was a scientific result that was later found to be 'mistaken' - I rarely claim to be 100% correct
ETA 5/8/20: I've been advised that the result was correct, it was the initial interpretation that needed to be withdrawn
Meta? I'd say so!
ETA 5/8/20: I've been advised that the result was correct, it was the initial interpretation that needed to be withdrawn
Meta? I'd say so!
Re: Astronomy question
Escape velocity as calculated from the (Earth's) surface is such a misleading concept.
I like to think of escape velocity as the velocity an object without an engine would need to escape Earth's (or other body's) gravity if it were fired from a cannon directly upwards (i.e. away from the Earth's centre of mass). The escape velocity is often calculated from the surface, but that does not mean that an interplanetary rocket would ever need to reach that velocity - it can accelerate until its increasing speed exceeds the escape velocity calculated for the height it reached, and only then cut the engine.
I like to think of escape velocity as the velocity an object without an engine would need to escape Earth's (or other body's) gravity if it were fired from a cannon directly upwards (i.e. away from the Earth's centre of mass). The escape velocity is often calculated from the surface, but that does not mean that an interplanetary rocket would ever need to reach that velocity - it can accelerate until its increasing speed exceeds the escape velocity calculated for the height it reached, and only then cut the engine.
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