# Solar radiation force on comets.

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1. Aug 7, 2014

### litup

I see this comet being chased by ESA, they said it is presently about 500 million km from the sun.

That puts the radiation received by that comet at about 10% (roughly) as what we receive on Earth, well at least on top of the atmosphere, at 1355 watts per meter ^2.

Looking at the size of that comet, I just rounded it out to 3000 by 4000 meters that would receive energy from the sun and at ten %, 135 watts per meter squared comes out to something like 1.5 gigawatts deposited on the surface at any given time, which of course goes up as it gets closer to the sun.

My question is, given 1.5 gw hitting the comet, how much force does that represent on the comet trying to push it away from the sun and how much would the orbit be altered by that force each orbit?

Could you calculate how much different the obit would be if it was orbiting say a black hole that doesn't radiate so there is no direct energy being deposited and it had a similar orbit to that comet.

That force would be trying to accelerate a mass about that of Mount Everest so I know it wouldn't have a massive effect but how much would that force move the orbit from orbit to orbit?

And wouldn't the orbit also be effected by the propulsion given by the ejected water when that water acts like a rocket leaving the comet, but in this case it could be aimed at any arbitrary angle.

So could you calculate all those potential changes to the orbit of that or any other comet?

2. Aug 7, 2014

### SteamKing

Staff Emeritus

The pressure is basically the energy flux divided by c (the speed of light) and multiplied by the cosine of the angle of incidence of the radiation to the illuminated surface.

The problem with an irregularly shaped object like a comet is it makes a simple calculation relatively complex, to account for the changing angles of incidence over the surface of the comet. (If the comet is tumbling as it orbits the sun, the calculation becomes exponentially more complicated).

3. Aug 7, 2014

### jbriggs444

The direct force of light pressure on the comet is quite small. That force can be derived by dividing the power (1.5 gigawatts by your estimate) by the speed of light to get a rate of momentum change. This follows from the equation e = pc that applies for photon energy.

1.5 gigawatts / c
= 1,500,000,000 kg m2/sec3 / 300,000,000 m/sec
= 5 kg m/sec2
= 5 Newtons

The force produced by the out-gassing water will be higher. It is more difficult to calculate but can be obtained by dividing the power carried away as kinetic energy in the outgassing water by the velocity of that outgassing water and multiplying by 2. This follows from the equation e = 1/2 mv2 (= 1/2 pv) that applies for kinetic energy at non-relativistic speeds.

Edit: The above estimate for force from incident ratiation assumes that the radiation is absorbed rather than being reflected and ignores any net force from re-radiation of lower energy photons.

Last edited: Aug 7, 2014
4. Aug 7, 2014

### litup

5 newtons. I guess it might have an effect in a couple million years:) So 5 newtons pushing against Mt Everest floating in space, 1.1 pounds, 2.2 kg, You'd have to really hang on with that much acceleration:)

Not much bang for the buck energy wise, eh.

5. Aug 7, 2014

### davenn

and probably nowhere near that much...

don't forget that the whole surface wont be facing the sun, lots, more than 1/2 of the surface, depending on its shape will; be in the shade.

so maybe only ~ 40% of that 1.5 GW

Dave

Last edited: Aug 7, 2014
6. Aug 7, 2014

### Orodruin

Staff Emeritus
You do not really need to start fiddling with cosines and surface orientations. The thing of importance is the cross-sectional area facing the sun. Given the approximate length scale of the comment, this should be enough to give a rough estimate of the order of magnitude of the effect as done above by jbriggs.