Deorbitization force.

[sao123]

How much force would it take to cause a projectile to fall out of orbit?
What angle would the force have to act?

For more clarity...even though highly improbable of being accomplished...

If you could hit a golf ball in space from a platform in low earth orbit, towards the planet... how far(hard) would you have to hit it for it to immediately lose orbit and fall to the earth. (Assuming it was a titanium ball and could withstand the heat and friction of atmospheric descent.) How fast would it be going when it hit earth? Would you hit it straight 90 degrees down toward the earth or would you hit at 60 degrees down toward the earth and 30 degrees opposit your orbit path?

 

[Fencer128]

Hi,

A quick couple of points. If you are talking about hitting a golf ball:

Hitting it with the club will cause the club to transfer momentum to the golf ball. The golf ball will accelerate whilst in contact with the club (and thus being acted on by an external force). Once the ball has left the club it will travel with constant velocity until acted upon by another force (i.e. friction due to the atmosphere).

So, the amount of momentum you need to transfer to the ball to make it fall to earth depends on how quickly you want it to do so. Even a small amount will cause it to move gradually towards the earth and eventually impact.

I'll leave the explanation for terminal velocity to someone else as this is my lunch break. I'm not sure about how the angle would effect the trajectory/velocity.

Good luck,

Andy

 

[Fencer128]

Another quick thought,

If you imagine the orbit to be a circle. If the angle you make to the tangent to the orbit means the ball will move inside the orbit, then it will eventually fall to earth. If the ball would move outside the orbit it may move to a wider orbit - or escape completely depending on the initial force applied. You can work all this out by balancing the centripetal and gravitational forces.

i.e for a stable orbit.

(GmM)/r^2 = -(mv^2)/r

where:

G is the gravitational constant
m is the mass of the ball
M is the mass of the earth
v is the velocity of the ball
r is the distance from the ball to the centrre of the earth

Good luck!

Andy

 

[sao123]

As I understand it, there is a specific minimum altitude an object must have at a certain velocity in order to be maintained in orbit, even if it is a very low and eventually decaying one. There is an inverse relationship between orbit altitude and velocity. At lower altitude, higher velocity is necessary to maintain orbit. At higher altitude, lower velocity is necessary to maintain the orbit.
Therefore to fall, the ball must cross below this threshhold of orbit altitude vs velocity. So I could either decrease the orbit altitude by some distance X, or decrease the velocity V. Or both, Right?

 

[Fencer128]

Hi,

As I understand it, there is a specific minimum altitude an object must have at a certain velocity in order to be maintained in orbit, even if it is a very low and eventually decaying one

Not a specific minimum altitude, just a specific altitude. You change the velocity you change the orbit. The comment about low and decaying orbits will just confuse you. This is due to friction with the atmosphere. In this example we're keeping things simple - so no atmospheric friction, no decaying orbits.

There is an inverse relationship between orbit altitude and velocity. At lower altitude, higher velocity is necessary to maintain orbit. At higher altitude, lower velocity is necessary to maintain the orbit.

Fair enough, you can see that from the equations too.

Therefore to fall, the ball must cross below this threshhold of orbit altitude vs velocity. So I could either decrease the orbit altitude by some distance X, or decrease the velocity V. Or both, Right?

If you move to a faster velocity then you move closer to the earth. This you have already explained above. If the force (or a component thereof) of the force acting on your ball accelerates it towards the earth then its orbital radius will decrease and its velocity will increase, as stated above.

I think I'm right here and I hope that's helped answer your question.

Andy

 

[Armitage]

Originally posted by: sao123
How much force would it take to cause a projectile to fall out of orbit?
What angle would the force have to act?

For more clarity...even though highly improbable of being accomplished...

If you could hit a golf ball in space from a platform in low earth orbit, towards the planet... how far(hard) would you have to hit it for it to immediately lose orbit and fall to the earth. (Assuming it was a titanium ball and could withstand the heat and friction of atmospheric descent.) How fast would it be going when it hit earth? Would you hit it straight 90 degrees down toward the earth or would you hit at 60 degrees down toward the earth and 30 degrees opposit your orbit path?

The speed of an object in a circular orbit is given by

Vc = sqrt(mu/r)

mu: gravitational parameter = 3.986012e5 km^3/s^2 for earth
r: radius of orbit (semi-major axis)

So, for an orbit with a 7000 Km radius

Vc = sqrt(3.986012e5 / 7000)
Vc = 7.546 Km/s

This is how hard you would have to hit your golf ball for it to fall "straight down" as you say.

The minimum dV required to have it simply hit the earth, you want to perform what's called a Hohmann Transfer. You are basically burning in plane such that you change your circular orbit to an elliptical orbit with a perigee radius equal to the radius of the earth. Actually, you could set your perigee radius higher due to the atmosphere and let drag pull it in the rest of the way. That's betond the scope of this post, and is left as an exercise for the reader

First find the energy of the required transfer orbit:

E = -mu/(r1 + r2) = -3.986012e5 / (7000 + 6378.135)
E = -29.795 Km^2/s^2

To find the velocity of the transfer orbit @ apogee

Va = sqrt(2*(mu/Ra + E))
Va = sqrt(2*(mu/7000 - 29.795))
Va = 7.369 Km/s

so the dV required is just the difference between these two velocities

dV = Vc - Va = 7.546 - 7.369
dV = 0.177 Km/s