Amateur question about classical physics - escape velocity

[MJJ]

As I understand it, escape velocity refers to the speed necessary for an object to escape the gravity of a body. Could the speed be lower if the object takes a less direct (<90-degree) trajectory, such as a plane uses? If so, then does escape velocity only apply to a vertical takeoff and trajectory?

 

Thanks.

Edited April 8, 2017 by MJJ

[swansont]

In the ideal case, it doesn't matter. The escape velocity is dependent on how much energy per unit mass you need to get arbitrarily far away from a planet (or other object that has appreciable gravity). It assumes no propulsion is involved and ignores complications such as air resistance.

 

It also assumes no motion once you are far away. Vertical motion addresses that by default. If you have any transverse motion, that costs extra.

[Country Boy]

If you assume atmosphere and that you craft has wings like an airplane, then you can use that lift to get higher. Then calculate the "escape velocity" from that higher level.

[OldChemE]

Perhaps a different way of looking at it will help you (a little College Physics)

 

Imagine you launch an object vertically with some starting velocity (tossing a ball in the air). Eventually, it reaches its maximum height and the velocity goes to zero, but the ball now has potential energy due to its height. We find escape velocity by doing this calculation assuming the height we reach is infinity. At infinity, potential energy goes to zero (no longer attracted to the earth), and kinetic energy goes to zero (zero velocity). The equation is 1/2 mv^2 - GMm/R = 0, where m is the mass of the object, M is the mass of the earth, R is the radius of the earth, G is the gravitational constant and v is the starting velocity (escape velocity). You Solve this equation for the value of v and that is the escape velocity. This is conservation of energy and the angle of travel has nothing to do with it.

Edited April 8, 2017 by OldChemE

[Bender]

If you assume atmosphere and that you craft has wings like an airplane, then you can use that lift to get higher. Then calculate the "escape velocity" from that higher level.

Since the atmosphere is rather thin, this would hardly have any effect on escape velocity.

[KipIngram]

Perhaps a different way of looking at it will help you (a little College Physics)

 

Imagine you launch an object vertically with some starting velocity (tossing a ball in the air). Eventually, it reaches its maximum height and the velocity goes to zero, but the ball now has potential energy due to its height. We find escape velocity by doing this calculation assuming the height we reach is infinity. At infinity, potential energy goes to zero (no longer attracted to the earth), and kinetic energy goes to zero (zero velocity). The equation is 1/2 mv^2 - GMm/R = 0, where m is the mass of the object, M is the mass of the earth, R is the radius of the earth, G is the gravitational constant and v is the starting velocity (escape velocity). You Solve this equation for the value of v and that is the escape velocity. This is conservation of energy and the angle of travel has nothing to do with it.

 

I don't see how the potential energy could undergo a step change, so that really just says we're taking infinity as the reference point for potential energy, right? So everywhere closer to the planet potential energy would be negative? If that's the case then we really just say that escape velocity is that velocity which gives the object a total energy (kinetic plus potential) of zero (keeping swansont's point about transverse velocity in mind).

 

Also, regarding "no longer attracted to the earth," that's imprecise. No matter how far away the object is it's always attracted to the earth with a non-zero force. But escape velocity means that it will also always have a non-zero velocity, and that velocity will always be enough to keep the object from reversing directions. No matter how far away it gets, if you stopped its motion and then watched it would (very slowly at first) fall back to the earth.

[goldglow]

As I understand it, escape velocity refers to the speed necessary for an object to escape the gravity of a body. Could the speed be lower if the object takes a less direct (<90-degree) trajectory, such as a plane uses? If so, then does escape velocity only apply to a vertical takeoff and trajectory?

 

Thanks.

I think escape velocity is the speed needed to escape gravity without further propulsion. For example, once a rocket launched to the moon from earth has reached a speed of about 25,000mph, it can turn of it's engines and still carry on to the moon without using more propellant to get there, and so save on weight, cost etc. ( It would still need fuel , although less, to escape the moon's gravity on the way back.). The rocket could still, i think, get to the moon at 10.000mph, or any other speed, but it would have to keep it's engines on all the time and burn fuel all the way there, with all that implies, or it would eventually be dragged back into orbit around the earth if it did turn off it's engines at those lesser speeds.Please someone correct me if i'm wrong.

Edited April 21, 2017 by goldglow

[KipIngram]

Actually it still needs fuel not just to escape the moon's gravity on the way back but also to achieve the delta-v necessary to have its orbit intersect the Earth's atmosphere. Once there, atmospheric friction will do the rest of the braking. But if you snapped your fingers and the moon vanished, the ship would still be in lunar orbit. It has to slow down in order to fall closer to the Earth. Atmospheric drag factors aside, it takes just as much delta-v to get from orbit to Earth as it does to get from Earth to orbit.

[swansont]

I don't see how the potential energy could undergo a step change, so that really just says we're taking infinity as the reference point for potential energy, right? So everywhere closer to the planet potential energy would be negative? If that's the case then we really just say that escape velocity is that velocity which gives the object a total energy (kinetic plus potential) of zero (keeping swansont's point about transverse velocity in mind).

 

Also, regarding "no longer attracted to the earth," that's imprecise. No matter how far away the object is it's always attracted to the earth with a non-zero force. But escape velocity means that it will also always have a non-zero velocity, and that velocity will always be enough to keep the object from reversing directions. No matter how far away it gets, if you stopped its motion and then watched it would (very slowly at first) fall back to the earth.

There's no step change. PE is indeed taken as negative in these kinds of problems.

 

You only get v=0 for this scenario when r gets to infinity; the force of attraction goes to zero in that limit.

I think escape velocity is the speed needed to escape gravity without further propulsion. For example, once a rocket launched to the moon from earth has reached a speed of about 25,000mph, it can turn of it's engines and still carry on to the moon without using more propellant to get there, and so save on weight, cost etc. ( It would still need fuel , although less, to escape the moon's gravity on the way back.). The rocket could still, i think, get to the moon at 10.000mph, or any other speed, but it would have to keep it's engines on all the time and burn fuel all the way there, with all that implies, or it would eventually be dragged back into orbit around the earth if it did turn off it's engines at those lesser speeds.Please someone correct me if i'm wrong.

The escape velocity changes with altitude, so you could have a rocket that fired its engines and cut them off before it gets to 25,000 mph and still have escape velocity for the altitude at cutoff.

[goldglow]

 

 

 

The escape velocity changes with altitude, so you could have a rocket that fired its engines and cut them off before it gets to 25,000 mph and still have escape velocity for the altitude at cutoff.

Good point- i didn't know that. Is that why the Apollo spacecraft went into earth orbit for a while instead of heading straight for the moon?

[swansont]

Good point- i didn't know that. Is that why the Apollo spacecraft went into earth orbit for a while instead of heading straight for the moon?

I'm not sure, but I imagine it was so that the launch was not constrained to happen at a specific time, and so that systems could be checked and confirmed to be OK after launch. You can do a part of an orbit before the lunar insertion burn, but if something went wrong immediately after launch, you could abort quickly. Once you head toward the moon, it cost a lot of fuel or several days to return. It's probably easier to time the insertion burn than the whole launch, as the latter has many more systems involved that might cause a delay.

[goldglow]

I'm not sure, but I imagine it was so that the launch was not constrained to happen at a specific time, and so that systems could be checked and confirmed to be OK after launch. You can do a part of an orbit before the lunar insertion burn, but if something went wrong immediately after launch, you could abort quickly. Once you head toward the moon, it cost a lot of fuel or several days to return. It's probably easier to time the insertion burn than the whole launch, as the latter has many more systems involved that might cause a delay.

Yes, that would be a very wise course of action - maybe get a boost from slingshot too. Thanks.

[KipIngram]

I'm not 100% sure about lunar, but interplanetary missions at least use Hohmann transfers. Imagine you're in a circular orbit around the central body. And you want to wind up in a different circular orbit. For lunar missions one of the orbits would be the moon's and the other a LEO; for planetary the two circular orbits would be those of the planet you're starting from and the planet you're going to.

 

There's an elliptical orbit that has the inner of those two circular orbit radii as its perigee / perihelion, and the outer as its apogee / aphelion. So a Hohmann transfer involves a burn that transfers you from your starting circular orbit into that elliptical orbit, and when you reach the radius of the other circular orbit you do another burn. For transfers outward both burns would increase your orbital speed; for transfers inward both would reduce your orbital speed. This is the lowest cost way to get between the two circular orbits. Obviously you have to time this so that the target body will "be there" when you arrive. This is why deep space mission have "launch windows."

 

I guess it's also worth noting that this is not the *fastest* way to get around. It's just the lowest energy cost way. As far as lunar missions go, this link has more:

 

https://en.wikipedia.org/wiki/Trans-lunar_injection

 

As far as "slingshot" goes, for a Hohmann transfer you would always go in the same direction your target is orbiting; otherwise you'd have to reverse direction when you arrived instead of just "speed up / slow down." But the precise timing would just be based on arriving at a "meeting" with the target; all you're really doing is shifting from circle to ellipse or ellipse to circle. I'm not an expert, but I'm guessing slingshots are done with bodies you encounter in the middle of a journey, to adjust direction or increase or decrease orbital speed wrt the sun. More here:

 

https://en.wikipedia.org/wiki/Gravity_assist

 

A wisely chosen slingshot might save energy or time compared to a straight-up Hohmann from origin to destination. But I bet you'd still do a Hohmann from origin to the gravity assist body. I'm not familiar enough with it to know what tricks might come after that, though.

 

Edit

 

Turns out that in some cases there's a bi-elliptic transfer that requires less energy than a direct Hohmann transfer:

 

https://en.wikipedia.org/wiki/Bi-elliptic_transfer

 

But those have really long time requirements.

Edited April 22, 2017 by KipIngram

[goldglow]

Thanks,K. Very helpful.