View Full Version : less than escape velocity

stitt29

2009-Nov-30, 03:22 PM

Hi,

I want to know how the velocity (or speed) needed to put a satellite into orbit is worked out. I know that 11.2km/s is the escape velocity when firing from the surface of the earth. Between 7.8km/s and 11.2km/s will put the object into an elliptical orbit. (I do know that satellites are not put into orbit this way, I am just interested how to calculate it).

Also how is the (average)radius of the orbit calculated for firing at a specified speed. i.e. 8km/s, 9km/s etc.

Also for Jupiter firing at above 59.5km/s will escape Jupiter. But again for speeds less than this, that would put the object into an elliptical orbit. How is the speed worked out and what average radius would result from which speed?

Otherworldly

2009-Nov-30, 03:47 PM

Hi,

I want to know how the velocity (or speed) needed to put a satellite into orbit is worked out. I know that 11.2km/s is the escape velocity when firing from the surface of the earth. Between 7.8km/s and 11.2km/s will put the object into an elliptical orbit. (I do know that satellites are not put into orbit this way, I am just interested how to calculate it).

Also how is the (average)radius of the orbit calculated for firing at a specified speed. i.e. 8km/s, 9km/s etc.

Also for Jupiter firing at above 59.5km/s will escape Jupiter. But again for speeds less than this, that would put the object into an elliptical orbit. How is the speed worked out and what average radius would result from which speed?

Probably the easiest way to work these things out is by conservation of energy. The force of gravity is

F = G*m1*m2/x2

where G is the universal gravitational constant, m1 and m2 are the masses of the two objects (the planet and the rocket, in this case), and x is the distance between the two. Energy is force acting through a distance, so to remove an object that is x0 from the center of the planet to an infinite distance requires

<integral from x0 to +infinity of> G*m1*m2/x2 dx

which works out to G*m1*m2/x0. This is the amount of energy the object needs to escape the planet's gravity.

The kinetic energy of an object is (1/2)*m1*v2 where v is its velocity, and m1 is its mass. Escape velocity is the velocity where the kinetic energy is just enough to escape the planet's velocity, so the two expressions can be set equal to each other.

G*m1*m2/x0 = (1/2)*m1*v2

The escape velocity is the velocity that solves this equation. The m1 on each side cancels with the other, and after rearranging a bit, the answer is

v = sqrt(2*G*m2/x0)

So the escape velocity depends on the mass of the planet m2, and the distance the object is already from the center of the planet. (If it is sitting on the surface, this is the radius of the planet.)

stitt29

2009-Nov-30, 04:09 PM

I am looking for less than escape velocity, that puts the object into an orbit and then finding the radius of this orbit. Thanks though I know its related to what you said. what is x0? is that x^0, i.e.=1,

is it easy to work out what I want from what you've said?

IsaacKuo

2009-Nov-30, 04:39 PM

You can calculate escape velocity with the equation:

ve = sqrt(2GM/r)

where v is escape velocity, G is the gravitational constant, M is the mass of Earth, and r is the distance from the center of the Earth.

At any given altitude, circular orbital velocity is given by:

v = sqrt(GM/r)

Notice the difference between escape velocity and circular orbital velocity is a factor of sqrt(2).

Otherworldly

2009-Nov-30, 04:50 PM

I am looking for less than escape velocity, that puts the object into an orbit and then finding the radius of this orbit. Thanks though I know its related to what you said. what is x0? is that x^0, i.e.=1,

No, x0 was meant to be the initial distance from the center of the planet. If you are calculating escape velocity from the surface of a planet, then x0 would just be the radius of that planet.

is it easy to work out what I want from what you've said?

I think we can get the answer using similar principles, specifically, conservation of energy.

To be in a circular orbit, the velocity is

v = sqrt(G*m2/r)

where G is the gravitational constant, m2 is the mass of the planet, and r is the radius of the orbit. (If you need a derivation of this result, just ask!)

So to go from the surface of a planet into orbit, energy is needed to increase altitude from x0 to r. (x0 is radius of the planet, r is the radius of the orbit.) This amount of energy is

<integral from x0 to r of> G*m1*m2/x2 dx

which is G*m1*m2*(1/x0-1/r). So this is the amount of energy needed to increase altitude. But kinetic energy is also needed to be in orbit. The amount of kinetic energy is

(1/2)*m1*(sqrt(G*m2/r))2 = (1/2)*m1*G*m2/r

So the total energy requirement is the sum of the two, which is

(1/2)*m1*m2*G/r+G*m1*m2*(1/x0-1/r) = G*m1*m2*(1/x0-1/(2r))

The kinetic energy of an object at the surface that is moving at v is

(1/2)*m1*v2

So if we set these two equal and solve for v, we get

v = sqrt[G*m2*(2/x0-1/r)]

So if you start at altitude x0 (the radius of the planet, if you are starting at the surface) and want to reach a circular orbit with radius r, the velocity needed is the expression above, where G is the gravitational constant, m2 is the mass of the planet, x0 is the radius of the planet, and r is the radius of the desired orbit.

There are several complicating factors. First, if the planet is rotating, the object on the surface already has some kinetic energy due to this rotation, and this energy is ignored in the above calculations. Second, there is atmospheric drag (if the planet has an atmosphere), so the true required velocity should be higher. Third, we can think about non-circular orbits, then the answer depends on how exactly you define the "average" radius of the orbit. Fourth, a rocket taking off does not leave the surface with escape velocity, but rises with a lower velocity, and continues to gain energy through the continued operation of the motors - but even if it leaves the planet, it never has the surface escape velocity.

Otherworldly

2009-Nov-30, 04:51 PM

Just to clarify, the expression from IsaacKuo (a similar expression is in my post) of v=sqrt(Gm/r) gives the velocity of an object once it is in orbit - not the velocity needed from the surface in order to achieve this orbit.

stitt29

2009-Dec-02, 01:25 PM

thanks, i worked out escape velocity for earth using the formula you gave. I might be a while working out the rest. Will post again

stitt29

2009-Dec-03, 01:30 PM

thanks otherwordly, i worked out putting an object into orbit for the earth and Jupiter, Quite easily from the formulas you gave me. great just what I wanted to know. Just one more little thing how is the minimum speed worked out? i.e. less than this the object returns to earth.

cjameshuff

2009-Dec-03, 07:37 PM

thanks otherwordly, i worked out putting an object into orbit for the earth and Jupiter, Quite easily from the formulas you gave me. great just what I wanted to know. Just one more little thing how is the minimum speed worked out? i.e. less than this the object returns to earth.

Earth has a non-zero radius. Disregarding atmosphere, if it hits the ground, it's returned.

There's no special limit below which orbits are impossible, the only thing keeping you from orbiting right now is that the ground is in the way. If Earth suddenly compacted into a grape-sized black hole, you would fall into an extremely elliptical orbit around it, reaching apogee at 1 old-Earth-radius with an orbital velocity equal to your current 1 old-Earth-circumference/day (464 m/s), and a perigee of about 11 km. (at which point you would be moving much faster, some 270 km/s)

Otherworldly

2009-Dec-03, 08:09 PM

sitt29, as cjamesstuff says, the constraint is you need to keep above the ground (or really, the atmosphere, which will slow you down and eventually cause you to hit the ground). So you just need to apply the same formula

v = sqrt[G*m2*(2/x0-1/r)]

with x0 being the radius of the earth (or other planet - x0 is where you are starting from) and r is a radius a little greater (just enough to get you above the atmosphere). Any velocity less than this, and you don't have enough energy to get up above the atmosphere, and stay there in orbit.

Same caveats as before - an object on the surface of a planet already has some energy if the planet is rotating, there is atmospheric drag, etc. But there is one other problem, which is that an object leaving the surface of earth even at the velocity specified by the formula (ignoring atmospheric drag) will not enter a circular orbit - rather, it enters an elliptical orbit, so the object will crash into the surface of the earth before one orbit is completed. At some point in its trajectory, the object will need to change direction in order to avoid this fate, so that requires additional energy, not included in derivation of the formula above.

For earth, if you start from the surface, and want to get maybe 200 miles above the ground, then the energy needed to gain this amount of altitude is only about five percent of the energy needed to achieve the orbital velocity. So it doesn't really matter too much what value of r you choose, unless you choose one that is much larger than x0. Then the amount of energy needed (and therefore the initial velocity) increases a lot.

Hop_David

2009-Dec-04, 08:27 AM

Earth has a non-zero radius. Disregarding atmosphere, if it hits the ground, it's returned.

There's no special limit below which orbits are impossible, the only thing keeping you from orbiting right now is that the ground is in the way. If Earth suddenly compacted into a grape-sized black hole, you would fall into an extremely elliptical orbit around it, reaching apogee at 1 old-Earth-radius with an orbital velocity equal to your current 1 old-Earth-circumference/day (464 m/s), and a perigee of about 11 km. (at which point you would be moving much faster, some 270 km/s)

I've heard a number of high school physics teachers say the path of a tossed penny is a parabola (disregarding air resistance). I used to try to correct them: "No it's a long skinny ellipse. The eccentricity is almost 1, but not quite. One focus is right in front of you, the other is earth's center." But to date none of them have paid attention. I guess the parabolic penny toss is one of those erroneous memes that won't die.

Hop_David

2009-Dec-04, 08:37 AM

Earth has a non-zero radius. Disregarding atmosphere, if it hits the ground, it's returned.

There's no special limit below which orbits are impossible, the only thing keeping you from orbiting right now is that the ground is in the way. If Earth suddenly compacted into a grape-sized black hole, you would fall into an extremely elliptical orbit around it, reaching apogee at 1 old-Earth-radius with an orbital velocity equal to your current 1 old-Earth-circumference/day (464 m/s), and a perigee of about 11 km. (at which point you would be moving much faster, some 270 km/s)

If you launched a rocket from earth's surface at a speed lower than escape velocity, it'll return to earth's surface.

You have to lift the rocket to an altitude above the atmosphere and then do the major rocket burn. If the burn's horizontal, the altitude of the burn will be your perigee altitude. (perigee is the the closest point to earth in an earth orbit)

I've heard a number of high school physics teachers say the path of a tossed penny is a parabola (disregarding air resistance). I used to try to correct them: "No it's a long skinny ellipse. The eccentricity is almost 1, but not quite. One focus is right in front of you, the other is earth's center." But to date none of them have paid attention. I guess the parabolic penny toss is one of those erroneous memes that won't die.

True, but it is effectively a parabola for any speed that a penny is likely to be traveling.

Otherworldly

2009-Dec-08, 05:16 AM

True, but it is effectively a parabola for any speed that a penny is likely to be traveling.

Yea, if we worry about a non-constant gravitational acceleration vector, why not go all the way and worry about relativistic effects :)

Hop_David

2009-Dec-08, 02:26 PM

Yea, if we worry about a non-constant gravitational acceleration vector, why not go all the way and worry about relativistic effects :)

Okay, okay, maybe I am being anal.

But this habit has been extended to larger tosses than a penny toss. For example, sub-orbital hops are sometimes called parabolic. For suborbital hops, ellipse eccentrity might differ from 1 enough that modeling it as a parabola will cause noticeable errors. It's a sloppy term.

Otherworldly

2009-Dec-08, 03:28 PM

But this habit has been extended to larger tosses than a penny toss. For example, sub-orbital hops are sometimes called parabolic. For suborbital hops, ellipse eccentrity might differ from 1 enough that modeling it as a parabola will cause noticeable errors. It's a sloppy term.

Without doing the maths, I suspect the misprediction of the landing spot probably would be noticeable :)

It would also differ from a parabola, because at some point, the engine turns off :)

mugaliens

2009-Dec-09, 02:02 AM

Hi,

I want to know how the velocity (or speed) needed to put a satellite into orbit is worked out. I know that 11.2km/s is the escape velocity when firing from the surface of the earth. Between 7.8km/s and 11.2km/s will put the object into an elliptical orbit. (I do know that satellites are not put into orbit this way, I am just interested how to calculate it).

Ug = -G * m1 * m2 / r

So,

Ve = sqrt ( 2 * G * M / r )

Which can be reduced to

Ve = sqrt ( 2 * g * r )

where:

g is the gravitational acceleration at r distance from the Earth's center of mass.

Also how is the (average)radius of the orbit calculated for firing at a specified speed. i.e. 8km/s, 9km/s etc.

Same equation, just reworked a bit:

r = Ve^2 / ( 2 * g )

Also for Jupiter firing at above 59.5km/s will escape Jupiter. But again for speeds less than this, that would put the object into an elliptical orbit. How is the speed worked out and what average radius would result from which speed?

Well, now that you have the tools... :)

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