- #1

Inquiring_Mike

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Approximately how much fuel would be necessary to place a satellite into geosynchronous orbit?

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- Thread starter Inquiring_Mike
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- #1

Inquiring_Mike

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Approximately how much fuel would be necessary to place a satellite into geosynchronous orbit?

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NateTG

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That depends on (among other things) the specific impulse of the fuel, and the mass of the satelite.

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Inquiring_Mike

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NateTG

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You also need to account for kinetic energy.

Once something is in orbit, no new energy is needed.

Once something is in orbit, no new energy is needed.

- #5

Ambitwistor

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Originally posted by Inquiring_Mike

Okay... Say the mass is 1000kg... Would using the W(total) = GMm (1/r - 1/r2) equation give the amount of energy needed to place the satellite into orbit...

That would give the amount of work just to raise it to the right altitude (i.e., to supply the necessary change in gravitational potential energy). But you'd need more energy to make it orbit circularly. By the virial theorem, the kinetic energy you'd have to add is equal to half of the final potential energy.

(By the way, I'm neglecting any kinetic energy the satellite might have started out with due to the Earth's rotation.)

- #6

enigma

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delta V vs. fuel can be determined from the ideal rocket equation:

[tex] \Delta V = -I_{sp}g_0 \ln{\frac{M_F}{M_I}} [/tex]

the orbital velocity of a circular orbit is related to distance from the center of the Earth by:

[tex] V = \sqrt{\frac{\mu}{r}} [/tex]

Where [tex]\mu[/tex] is the gravitational parameter and is 398,600 km^3/sec^2 for Earth.

In low Earth orbit, roughly 200km (=6578km to center of the Earth), the orbital velocity is ~7.8km/sec

In geo-synch (6.6 Earth Radii ~= 42100 km), the orbital velocity is ~=3.1 km/sec

The most energy efficient way to transfer to another orbit is typically a http://liftoff.msfc.nasa.gov/academy/rocket_sci/satellites/hohmann.html

The semimajor axis of the transfer ellipse is [tex]\frac{r_{LEO} + r_{GEO}}{2}[/tex], ~24250km

The velocity at any point on the ellipse (the above is merely a special case where a = r) is:

[tex]V = \sqrt{\mu(\frac{2}{r}-\frac{1}{a})}[/tex]

The velocity in LEO to get into the transfer ellipse (with LEO for r and trans for a) works out to ~10.4 km/sec

The velocity of the transfer ellipse when it gets to GEO (GEO for r and trans for a) is ~1.6km/sec

The delta V needed is obtained by finding the magnitude of the change in each spot.

3.1 - 1.6 + 10.4 - 7.8 km/sec = 4.1 km/sec delta V

That delta V is then plugged into the ideal rocket equation. With a reasonable specific impulse (Isp) of 250 for a monopropellant rocket (which are typically used on satellites), and your spacecraft mass of 1000kg:

[tex] 4100\frac{m}{s} = -250*9.8*\ln{\frac{1000}{1000+fuel}} [/tex]

Gives 4330kg of fuel required for the transfer maneuver. That is on top of the ~ 9km/sec it takes to get into low Earth orbit (7.8km/sec required + drag losses during ascent)

Judging from the mass of fuel needed, it stands to reason that you'd use a more powerful launch vehicle with a bi-propellant upper stage to get the satellite to GEO, which is exactly what they do.

[tex] \Delta V = -I_{sp}g_0 \ln{\frac{M_F}{M_I}} [/tex]

the orbital velocity of a circular orbit is related to distance from the center of the Earth by:

[tex] V = \sqrt{\frac{\mu}{r}} [/tex]

Where [tex]\mu[/tex] is the gravitational parameter and is 398,600 km^3/sec^2 for Earth.

In low Earth orbit, roughly 200km (=6578km to center of the Earth), the orbital velocity is ~7.8km/sec

In geo-synch (6.6 Earth Radii ~= 42100 km), the orbital velocity is ~=3.1 km/sec

The most energy efficient way to transfer to another orbit is typically a http://liftoff.msfc.nasa.gov/academy/rocket_sci/satellites/hohmann.html

The semimajor axis of the transfer ellipse is [tex]\frac{r_{LEO} + r_{GEO}}{2}[/tex], ~24250km

The velocity at any point on the ellipse (the above is merely a special case where a = r) is:

[tex]V = \sqrt{\mu(\frac{2}{r}-\frac{1}{a})}[/tex]

The velocity in LEO to get into the transfer ellipse (with LEO for r and trans for a) works out to ~10.4 km/sec

The velocity of the transfer ellipse when it gets to GEO (GEO for r and trans for a) is ~1.6km/sec

The delta V needed is obtained by finding the magnitude of the change in each spot.

3.1 - 1.6 + 10.4 - 7.8 km/sec = 4.1 km/sec delta V

That delta V is then plugged into the ideal rocket equation. With a reasonable specific impulse (Isp) of 250 for a monopropellant rocket (which are typically used on satellites), and your spacecraft mass of 1000kg:

[tex] 4100\frac{m}{s} = -250*9.8*\ln{\frac{1000}{1000+fuel}} [/tex]

Gives 4330kg of fuel required for the transfer maneuver. That is on top of the ~ 9km/sec it takes to get into low Earth orbit (7.8km/sec required + drag losses during ascent)

Judging from the mass of fuel needed, it stands to reason that you'd use a more powerful launch vehicle with a bi-propellant upper stage to get the satellite to GEO, which is exactly what they do.

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- #7

enigma

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Originally posted by Inquiring_Mike

Then would the satellite need more energy to keep it in orbit?

Atmospheric drag is totally insignificant in GEO, even if your sat is built like a parachute. In LEO, you need to re-boost occasionally to counter drag, but in GEO, the only fuel you'd need would be for stationkeeping (keeping it in the right position in the sky). That's usually up to a few tens of m/s per year, depending on the specific placement of the orbit.

- #8

Nereid

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... and its atmosphere.

A GEOsat experiences the gravitational effects of the Moon, Sun, the other planets (esp Jupiter) ... in a real solar system, a geostationary orbit is not stable ... what is the characteristic time for its decay? What proportion of sats would (eventually) crash into the Moon? the Earth? leave the Earth-Moon system entirely?

- #9

enigma

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Variations in the Earth's gravitational field due to non-symmetry have orders of magnitude greater effect than most other perturbation sources other than low altitude atmospheric drag.

To escape the Earth/Moon system from GEO, you need a delta V of ~2.75km/sec. That works out to an impulsive burn of over 140% of the mass of the spacecraft in fuel. A perturbation will not give that sort of energy.

To crash into the moon, you need to increase the velocity by 2.3km/sec. What's more, that change needs to be all in the in-track direction. The only perturbation which constantly acts in a single track-wise direction is atmospheric drag. Luni- and helio- gravitational effects act to pull the sat along the orbit when it is getting aligned with the body and against the orbit when it's going away. All they do is 'wobble' the orbit.

No satellite will crash into the moon unless it is hit by a big meteoroid. They will eventually degrade into the Earth, but you'll have a REALLY long wait from GEO. The atmosphere is a bit thin.

- #10

Nereid

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IIRC, the Jovian rings are essentially permanent, though the individual ring particles not. The spectacular rings of Saturn are thought to be far from permanent, with a decay time of the order of 100 My - what causes the main ring (A, B, C) particles to decay (the E ring - and others? - are like the Jovian ones, constantly replenished)? What's the story for Uranus and Neptune?

Of course, 100 My is a REALLY, REALLY long time for good ole homo sap., but there've been ~45 of these periods since the solar system formed.

What is the characteristic decay time of GEOsats?

For small GEOsats, how important are electrostatic and magnetic effects? E.g. non-gravitational forces due to build up of charge and resultant interaction with magnetic fields.

- #11

Golfer

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That depends on (among other things) the specific impulse of the fuel, and the mass of the satelite.

This is a correct statement. However, a convient rule of thumb is that between 1 - 2 % of the total mass of the launch system [satellite, launch vehincle, propellant] can be plaaced in orbit.

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