View Full Version : Watts, Newtons, help me get the units right!

2012-Jun-08, 10:57 AM
While watching the Venus transit, I was thinking "What would it take to move the entire planet Venus so it could break away from the straight path across the Sun, would even the entire power of the Sun be enough?"

So, more formally: If the entire power output of the Sun (some 4e26 Watts, I think) was put to work accelerating the planet Venus instead of producing photons, neutrinos and helium, what acceleration could be provided?

Intuitively, Newton's formula F = m*a seems to be the one I need. "a" is the quantity I am seeking, "m" is the mass of Venus which is easy to look up. So, if I only had "F" I could compute "a". "F" is the force, in Newtons. But I don't have any Newtons to do the job, instead I have some number of Watts.
Watts = Joules / second.
Joules = Newton*meters.
So Watts is Newtons*meters/second
So, if I could divide those Watts with a velocity (meters/second) I'd have the required Newtons to put into the F = m*a (or rather a = F/m) formula. But what velocity would that be?
I've gotten the units mixed up somehow, can you please help?

2012-Jun-08, 11:16 AM
The problem you've hit is that energy alone can't accelerate anything, you need impulse for that.

Now, taking an ion thruster such as the NSTAR (2100 Watts for 92 millinewtons) and assuming it can be scaled linearly you get 1.6 1022 newtons which will accelerate Venus by 0.003 m/s2.
This while drinking 2.91014 tons of xenon per second as reaction mass.

And this is one of the most efficient uses of reaction mass.

2012-Jun-08, 11:43 AM
I guess you want to increase Venus's speed to escape velocity. Escape velocity is square root of two times orbital velocity.

2012-Jun-11, 05:21 PM
Remember Watts (W) is Energy / time not energy. What you realy need is the total energy required. (As noted above you would need to address momentum as well)

E(J) = P(W) * t(s) = m*v^2

So find out how much faster Venus would need to go to reach escape velocity and plug it into
time(seconds) = (mass of Venus * v^2)/(2*Suns Power output)

This would give you the time it takes form the Sun to output enough energy to accelerate Venus to escape velocity (49.5 Km/s according to Wikipedia but Venus also has an orbital; velocity of 35 Km/s you may be able to take advantage of).

2012-Jun-12, 04:41 AM
So plugging in the numbers, doubling Venus's kinetic energy by going from about a circular orbit to escape (3.5e4 to 4.95e4 m/s) would require 2.982e33 J. If we had some way of harnessing the sun's 3.846e26 J/s of power into losslessly increasing Venus's kinetic energy, we would need to apply 7.754e6 s (about 90 days) worth of the sun's power to Venus at perigee for it to leave the solar system (or less if we can send it out with a Jupiter slingshot).