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Calculating escape velocity
Hypatia
Registered User regular
Registered User regular
Let me preface this by saying that this isn't for a class or for anything where I need to exhibit my prowess at physics, this is a "just for fun" thing.
That being said, I am a total idiot when it comes to physics. I am trying to figure out what speed something would have to be traveling to break orbit/be at escape velocity if it were starting from a point that's about 4600 km from Earth.
I've been staring at Wikipedia pages and physics help pages for the last day or so, and it's like they're in a foreign language. Can anyone help me out?
That being said, I am a total idiot when it comes to physics. I am trying to figure out what speed something would have to be traveling to break orbit/be at escape velocity if it were starting from a point that's about 4600 km from Earth.
I've been staring at Wikipedia pages and physics help pages for the last day or so, and it's like they're in a foreign language. Can anyone help me out?
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Of course, an object doesn't need to be travelling at escape velocity, as long as it has some kind of thrust generator on it, it can move any speed.
1/2 mv^2 is correct for the kinetic energy of a body with mass m and velocity v, but potential energy due to gravity is GMm/r, for M the mass of one body, m the mass of the other body, G the gravitational constant, and r the distance between the bodies of mass M and m.
So, to find escape velocity:
(1/2)mv^2 = GMm/r
(1/2)v^2 = GM/r
v = sqrt(GM/2r)
The reason that you don't necessarily have to have a velocity equal to escape velocity while under thrust is fairly simple. 'Escape velocity' is defined as the velocity which you must have in order to travel from a given point (defined as r, the distance from the center of mass of the body which you are escaping) to infinite distance with no residual velocity. In other words: if you were in a universe composed of you and the Earth, and you left the surface of Earth at escape velocity, you would fly forever in your initial direction with no additional acceleration.
If you are capable of continuously accelerating, however, there's no need to have sufficient velocity to escape the Earth's gravitational potential forever. All you need to do is accelerate enough to maintain whatever distance from Earth you want. Which means that, if you want a net velocity of zero in the radial direction from Earth, you need an acceleration equal to the acceleration due to gravity at that distance from Earth.
If you don't want to continually accelerate, though, and instead want to get up to some altitude and then stay there a while without having to work anymore, you need enough kinetic energy to overcome the gravitational potential that would otherwise pull you back down. But you don't want to get any further form Earth, so you don't want to have your velocity vector pointing away from the surface. Instead you need to angle your velocity vector tangent to a circle around the Earth at your desired altitude. You'll be continuously accelerated by Earth's gravity, but your tangential velocity vector will just spin in a circle rather than being pulled around to point back at the Earth (ignoring drag and so forth).
So when a rocket is lifting off the surface, they aren't accelerating to escape velocity or continually accelerating until they get high enough and then stopping: they accelerate both away from and around Earth, into their desired orbit. Even a lunar flight doesn't actually have to reach Earth's escape velocity, necessarily; it just has to enter an orbit that intersects the moon's.
On the other hand, if you want to talk about a space craft that's leaving our solar system entirely, the escape velocity of Earth isn't sufficient. You'll have to either continue accelerating your way out of the system or achieve escape velocity from the system as a whole. You can calculate that latter velocity the same was as Earth's, only using the total mass of the solar system. Generally it's sufficiently accurate to just use the mass of the sun since it's so much greater than the masses of the planets (three orders of magnitude more than Jupiter, which is itself three orders of magnitude larger than Saturn).
I can't personally help you, but I know someone, or more accurately, something that can:
Kerbal Space Program
Enjoy!
Or is that too simple?
You've already that far away from earth without falling back into the gravity well, at least that's how I'm reading the problem. It has been TOO long since I've done any physics.
If you're already in a stable orbit 4600 km from earth's surface, your object is going ~6 km / s or 21000 km / h. That means you would need to go about 5.2 km / s (18720 km / h) faster.
Accelerating at 1 g would get you there in about 530 seconds, or slightly under 9 minutes. Since we're ignoring drag (reasonably, since we're in space) there is no rush to get up to velocity (and 1 g is usually a hell of a lot of sustained force), so adjust your figures as necessary.
csgnetwork.com/satorbdatacalc.html
Yup, sorry. Happens when you use wiki without thinking.
Just in case you're dismissing chrishallett's suggestion as a "ha ha" friendly joke, take it seriously! If you have an interest in real-life rocket science or orbital physics, Kerbal Space Program is one of those magical things that is both fun and educational. Playing it has given me not only a better knowledge of how rocket science works, but also a greater appreciation for the accomplishments of our real-life space programs.