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The moon is edging away from the earth. Does this mean that its velocity is slightly more than escape velocity?

Grumble wrote: Thu Jan 20, 2022 11:59 pm The moon is edging away from the earth. Does this mean that its velocity is slightly more than escape velocity?

They are related, but basically no.

The escape velocity is √2 times the orbital speed, so is always about 41% larger.

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.

Gfamily wrote: Fri Jan 21, 2022 12:17 am

Grumble wrote: Thu Jan 20, 2022 11:59 pm The moon is edging away from the earth. Does this mean that its velocity is slightly more than escape velocity?

They are related, but basically no.

The escape velocity is √2 times the orbital speed, so is always about 41% larger.

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/a^⅔ ~ a^⅓

Fat fingered the quote button instead of the edit button.

dyqik wrote: Fri Jan 21, 2022 12:25 am

Gfamily wrote: Fri Jan 21, 2022 12:17 am

Grumble wrote: Thu Jan 20, 2022 11:59 pm The moon is edging away from the earth. Does this mean that its velocity is slightly more than escape velocity?

They are related, but basically no.

The escape velocity is √2 times the orbital speed, so is always about 41% larger.

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/a^⅔ ~ a^⅓

Yes, but the escape velocity from a particular orbit also depends on the altitude.

Gfamily wrote: Fri Jan 21, 2022 12:29 am

dyqik wrote: Fri Jan 21, 2022 12:25 am

Gfamily wrote: Fri Jan 21, 2022 12:17 am

They are related, but basically no.

The escape velocity is √2 times the orbital speed, so is always about 41% larger.

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/a^⅔ ~ a^⅓

Yes, but the escape velocity from a particular orbit also depends on the altitude.

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.

dyqik wrote: Fri Jan 21, 2022 1:08 am

Gfamily wrote: Fri Jan 21, 2022 12:29 am

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/a^⅔ ~ a^⅓

Yes, but the escape velocity from a particular orbit also depends on the altitude.

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.

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.

Ignore

dyqik wrote: Fri Jan 21, 2022 12:25 am

Gfamily wrote: Fri Jan 21, 2022 12:17 am

Grumble wrote: Thu Jan 20, 2022 11:59 pm The moon is edging away from the earth. Does this mean that its velocity is slightly more than escape velocity?

They are related, but basically no.

The escape velocity is √2 times the orbital speed, so is always about 41% larger.

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/a^⅔ ~ a^⅓

My maths is way off here. Orbital velocity is proportional to 1/√a

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.

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.

Sciolus wrote: Fri Jan 21, 2022 9:45 am 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.

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.

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.)

dyqik wrote: Fri Jan 21, 2022 2:12 am 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.

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.

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.