Orbiting satellite coming back to land...

[Externet]

Hi.

 

As the shuttles, when returning to earth have to dissipate its enormous kinetic energy by counter propulsion (expensive) or braking trough the atmosphere; what if the orbit is geostationary ?

 

Would then a tiny impulse towards earth to start falling from orbit and then a parachute suffice ?

[Janus]

To go from geostationary orbit to one that would brush the atmosphere requires a delta v of about 1.5 km/sec. A tiny nudge would just put into an elliptical orbit with a slightly smaller period(it would no longer be geostationary) and with a perigee a little closer to Earth. It takes the 1.5 km/sec change to get that perigee to be within the Earth's atmosphere.

[J.C.MacSwell]

Hi.

 

As the shuttles, when returning to earth have to dissipate its enormous kinetic energy by counter propulsion (expensive) or braking trough the atmosphere; what if the orbit is geostationary ?

 

Would then a tiny impulse towards earth to start falling from orbit and then a parachute suffice ?

 

A satellite in geostationary orbit is moving quite fast relative to the Earth, any point on it, or it's atmosphere. (it has to, in order to "keep up" while following a much longer path)

 

The associated kinetic energy, along with the gravitational potential energy of it's height above the Earth, must be dealt with prior to any normal parachute being able to take over.

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To go from geostationary orbit to one that would brush the atmosphere requires a delta v of about 1.5 km/sec. A tiny nudge would just put into an elliptical orbit with a slightly smaller period(it would no longer be geostationary) and with a perigee a little closer to Earth. It takes the 1.5 km/sec change to get that perigee to be within the Earth's atmosphere.

 

This seems low?

[D H]

To go from geostationary orbit to one that would brush the atmosphere requires a delta v of about 1.5 km/sec. A tiny nudge would just put into an elliptical orbit with a slightly smaller period(it would no longer be geostationary) and with a perigee a little closer to Earth. It takes the 1.5 km/sec change to get that perigee to be within the Earth's atmosphere.[/quote']This seems low?

Janus is right. The vis-viva equation is

 

[math]v^2 = \mu\left(\frac 2 r - \frac 1 a\right)[/math]

 

The velocity of an object in a circular orbit at radius [math]a[/math] is thus

 

[math]v = \sqrt{\frac{\mu}{a}}[/math]

 

The delta-v needed to change a circular orbit at radius [math]a[/math] to an elliptical orbit with apogee [math]a[/math] and perigee [math]r[/math] is

 

[math]\Delta v = v\left(1-\sqrt{\frac{2r}{a+r}}\right)[/math]

 

For a geosynchronous satellite, a=42,164 km and v=3.0747 km/sec. You want to target an orbit with a perigee 60 km to 120 km above the surface of the Earth (R_E= 6378 km). Targeting a 60 km altitude requires a delta V of 1.492 km/sec; for a 120 km altitude this reduces to 1.486 km/sec. Both are 1.5 km/sec to two decimal places.

[Externet]

Unsure to understand...

 

If a geostationary satellite is wanted to land directly and vertically down on the equator, is there more than that 1.5km/sec initial push downwards needed to bring it down with a simple parachute, ? (discarding winds)

 

In reverse, could a geostationary satellite placement from a launch station on the equator be carried by only a vertical straight up 'flight' and release it with zero velocity when reaching the proper altitude ?

[insane_alien]

no.

 

to put a geostationary satellite up there you need to provide a massive horizontal component to its velocity. 3.07km/s to be precise although it would be going much faster during its ascent probably around 9km/s but depends on the trajectory.

 

a simple vertical ascent will not produce the necessary velocity.

 

to deorbit it, the thrust is applied in the opposite direction from its motion, not downwards.

[Externet]

:-(

Now am complicating myself...

I thought that the launch rocket had horizontally the same angular speed of the spinning earth surface, and if when rising vertically up from the equator, the satellite would carry such same 'horizontal' speed component when reaching the straight-up altitude, staying geosynchronous above the equatorial launch site because of that 'horizontal' speed, keeping the launch site directly underneath.

 

Sorry for the poor wording but have some difficulty expressing it properly.

 

Thanks.

[Cap'n Refsmmat]

Imagine spinning around in a circle while holding a gun. You shoot. Does the bullet keep going around in a circle with you (getting farther and farther away, of course), or does it just shoot straight out and hit whatever it was pointed at when you fired?

 

To get an object to move in a curved trajectory requires a force. Trace out the path of a satellite going to geosynchronous orbit as the earth rotates; it's curved.

[D H]

I thought that the launch rocket had horizontally the same angular speed of the spinning earth surface, ...

It does. Even if launched from the equator, that is a relatively small speed compared to the velocity needed to get into orbit:

 

[math]v=r\omega = 6378\,\text{km}\cdot\frac{2\pi}{23.9344696\,\text{hours}} = 465\,\text{m}/\text{sec}[/math]

 

Compared to the velocity of a vehicle low-Earth orbit (about 7785 m/sec), that 465 m/sec is pretty tiny.

 

... and if when rising vertically up from the equator, the satellite would carry such same 'horizontal' speed component when reaching the straight-up altitude, staying geosynchronous above the equatorial launch site because of that 'horizontal' speed, keeping the launch site directly underneath.

A geosynchronous satellite has a velocity of about 3,000 m/sec. That 465 m/sec? Not much. Moreover, launching vertically would be mind-boggling costly in terms of fuel. The gravity losses would be immense.

 

A rocket launches vertically not because that is the "best" launch angle. A rocket launches vertically because the rocket would fall over were it tilted at the "best" angle.

 

Suppose the vehicle is climbing. If it weren't thrusting, the flight path angle, the angle between the velocity vector and local horizontal, would decrease over time because gravity will reduce the vertical component of velocity. Because of drag, a non-thrusting rocket will have a zero angle of attack. Gravity and drag will turn the rocket. (Think of how an arrow flies.)

 

Now suppose the vehicle is thrusting. The best angle at any point along the atmospheric part of the ascent is a zero angle of attack. Gravity (and drag) will turn the vehicle so by the time burnout occurs the vehicle is (a) at orbital insertion altitude, (b) is flying more-or-less horizontally, and © has a velocity that is close to (or even exceeds) orbital velocity.

 

How does the satellite get to geosynchronous orbit? The payload of the main rocket is the satellite plus another rocket. That rocket will fire horizontally just long enough to make the orbit be an ellipse with apogee at geosynchronous altitude. The rocket+satellite then coast along this orbit. Ignoring correction burns, it doesn't fire again until it reaches geosynchronous altitude. Then it fires horizontally again to bring the vehicle up to geosynchronous velocity.

[J.C.MacSwell]

Janus is right. The vis-viva equation is

 

[math]v^2 = \mu\left(\frac 2 r - \frac 1 a\right)[/math]

 

The velocity of an object in a circular orbit at radius [math]a[/math] is thus

 

[math]v = \sqrt{\frac{\mu}{a}}[/math]

 

The delta-v needed to change a circular orbit at radius [math]a[/math] to an elliptical orbit with apogee [math]a[/math] and perigee [math]r[/math] is

 

[math]\Delta v = v\left(1-\sqrt{\frac{2r}{a+r}}\right)[/math]

 

For a geosynchronous satellite, a=42,164 km and v=3.0747 km/sec. You want to target an orbit with a perigee 60 km to 120 km above the surface of the Earth (R_E= 6378 km). Targeting a 60 km altitude requires a delta V of 1.492 km/sec; for a 120 km altitude this reduces to 1.486 km/sec. Both are 1.5 km/sec to two decimal places.

 

Thanks DH, now I see what Janus meant.

 

I was thinking a delta v around 2.5 km/s would be required to match the "speed" of the atmosphere at re-entry, getting it down around 0.5 km/sec and enabling a parachute to take over. This of course no longer being an orbit.

 

So in the above orbit, what is the speed at pergee/re-entry? Isn't it even higher than the original 3,0747 km/sec?

[D H]

I was thinking a delta v around 2.5 km/s would be required to match the "speed" of the atmosphere at re-entry, getting it down around 0.5 km/sec and enabling a parachute to take over. This of course no longer being an orbit.

 

That's how things work in (bad) science fiction movies. That is not how things work in the real world.

 

To reenter that way would require using the equivalent of a launch vehicle for reentry. The launch vehicle to launch that big honking reentry vehicle would be immense, orders of magnitude larger than the biggest launch vehicle ever created. If that is how we had to reenter, there would be no reentry. Every flight into space would be a one-way journey.

 

Think of it this way: Why use thrusters to bring the vehicle down to near zero velocity when the atmosphere does that for us for free? In fact, that Mars has an atmosphere that can do most of the work of slowing a lander but the Moon does not is why you will see claims that a trip to Mars is cheaper (fuel-wise) than is a trip to the Moon. Landing on the near airless Moon requires the use of thrusters all the way down. Vehicles that land on Mars can use some kind of aero-assist.

 

So in the above orbit, what is the speed at pergee/re-entry? Isn't it even higher than the original 3,0747 km/sec?

Much higher. It is much higher than low-Earth orbit velocity of about 7.8 km/sec. For something reentering from geosynchronous altitude, reentry velocity is about 11.8 km/sec. That's high (higher than the Apollo reentry; it non-intuitively takes more energy to get to/from GEO than it takes to get to/from low lunar orbit) but it is less than the reentry velocity experienced by the Stardust mission (12.4 km/sec).

[J.C.MacSwell]

 

That's how things work in (bad) science fiction movies. That is not how things work in the real world.

 

Thanks DH

 

That is what I thought. I think I now know what Janus meant by "brush the atmosphere" as well. I took it to mean matching the velocity of the atmosphere and ready for the parachute as suggested in the OP.

 

Obviously that would take a lot of energy to accomplish.