How much energy is needed to give Venus an infinite day length?
The rotational kenetic energy of anything is: K = I(Ω^2)/2 I = the moment of inertia Ω = the angular velocity I is really a complicated integration of the mass of the rotating item over its radius but a good approximation is: I = mr^2 Using the mass and radius of Venus: m = 4.86900 x 10^24 kilograms r = 6.0518 x 10^6 meters I = 1.7832 x 10^38 kg x m^2 For Ω, we must convert the time of one rotation from days to seconds while dividing that time into 2π. Ω = 2π/243 days/24 hours/60 min/60 seconds = 2.993 x 10^-7 rads x sec^-1 K = {1.783 x 10^38(2.993 x 10^-7)^2}/2 K = 7.984 x 10^24 Joules Assuming 100% transfer of energy in your design, the above amount of energy will stop Venus. BUT! You want to reverse the retrograde rotation of Venus so that its day is the same length as its year (224.7 Days).
The day and year are almost the same length now. If they were the exact same length, Venus would always keep the same side facing the sun, and that side would have an “infinite” day length. It would take 10 to the 29th power joules to speed up Venus to a 24-hour day. So it would take a lot less than that huge amount to speed it up just a little, so that its 243-earth-day day length could become equivalent to its 225-earth-day year length. I’m not math-sophisticated enough to work that out, but try contacting this person: davejanell@comcast.net I don’t know if that email address is still good, because his webpage was last updated in 2005.
Related Questions
- Imagine in the far future, if we could easily create mass from energy and then used this to seed as many planets as we wanted, only we want them in the goldilock zone. How many earth sized planets could orbit within it without crashing into each other?
- Exactly how far is Earth from the sun in km?
- Is it possible to terraform Venus?
