Geosynchronous orbital drop duration and velocity

[MKruer]

Can someone tell me if this is correct.

Geosynchronous orbit = 35,787,055m
Gravitational acceleration = 10m/s/s
2675s => 44.58m
26,750m/s @ Impact

[Hamiltonian]

It's difficult to tell what you're doing. Are you calculating how long it will take to fall from geosynch orbit?
If so, you're forgetting that gravity is lower at geosynch height, so the initial acceleration will be lower than 10m/s/s.
The calc for the speed of impact just needs a conversion of the potential energy difference to kinetic. But for the time to fall, you have to integrate inverse velocity over distance.

[Malenfant]

Is this through an atmosphere? If so, don't you need to account for terminal velocity?

[Hamiltonian]

Is this through an atmosphere? If so, don't you need to account for terminal velocity?

35700 km of accleration outside the atmosphere first, so something falling from GEO will hit air at about 10km/s. Will terminal velocity have time to get relevant? Small things will burn, big things will hit the ground supersonicly.

[Malenfant]

Is this through an atmosphere? If so, don't you need to account for terminal velocity?

35700 km of accleration outside the atmosphere first, so something falling from GEO will hit air at about 10km/s. Will terminal velocity have time to get relevant? Small things will burn, big things will hit the ground supersonicly.

True

[MKruer]

Sorry about that everyone. I was tired when I posted that and wasn?€™t very clear. I was trying to calculate the speed of an object hitting the ground if it were dropped from geostationary orbit. I already see I screwed up by not talking into account that the gravitational acceleration is different at 32,000km so the duration and final impact speed will be substantially less.

For the time being let?€™s calculated with out any atmospheric resistance, it will make a difference but only very late in the fall. The atmospheric resistant can be calculated later.

[selden]

Remember, things don't "drop". They travel on an elliptical orbit. It's just that its perigee is below the surface of the Earth.

[MKruer]

Ok then it has an elliptical orbit with and Eccentricity of 1 relative to the surface of the planet

its a hypothetical question.

[selden]

If you think of it as an orbit, though, it's easier to calculate the time it takes.

[Hamiltonian]

The change in potential energy between GEO and Earth's surface gives a speed of 10.3 km/s.
Integrating the straight drop over the same distance comes out to 14,830s. (4.1 hours).

Arthur C Clarke wrote a story called Jupiter V where he talked about working out the time for a vertical drop from orbit. If you treat it like there's a point mass in the middle, the time to get all the way to the centre is always 1/(4.root(2)) times the circular orbital period.

[tony873004]

This problem either requires calculus or an n-body simulation.

Using n-body with Gravity Simulator:

Ignoring atmosphere, it would take it 4 hours 7 minutes, 53 seconds to fall from geosychronous orbit, assuming that its orbital velocity was zero. It would be travelling 10.3 km/s as it hit the ground.

[Hamiltonian]

Ignoring atmosphere, it would take it 4 hours 7 minutes, 53 seconds to fall from geosychronous orbit, assuming that its orbital velocity was zero. It would be travelling 10.3 km/s as it hit the ground.

Checks my calculus nicely.
Thanks.

[tony873004]

lol. I didn't even see your post when I posted my sim numbers. How do you integrate the drop?

[Hamiltonian]

lol. I didn't even see your post when I posted my sim numbers. How do you integrate the drop?

The velocity just comes from the change in potential energy. It is root(2GM(R-r)/(Rr)), if R and r are your start and finish distances.
Because time is distance over velocity, you must integrate the inverse of the velocity w.r.t. distance, between the limits R and r.
It comes down to solving the integral of root(x/(a-x)), which is too messy to write down here.