As an object falls, its speed increases because it’s being pulled on by gravity. The acceleration of gravity near the earth is g = -9.81 m/s^2. To find out something’s speed (or velocity) after a certain amount of time, you just multiply the acceleration of gravity by the amount of time since it was let go of. So you get: velocity = -9.81 m/s^2 * time, or V = gt. The negative sign just means that the object is moving downwards. If it were positive, then it would be moving up. For speed rather than velocity, you just drop the negative sign.
If you have an initial velocity (if you threw the ball up or down instead of just letting go of it), then you have to include this in the equation, too, giving you: V = Vo + gt, where Vo is the initial velocity of the object. This equation will still work if you threw the ball to the side, instead of straight up or down, except that it will only give you the up-down velocity, not the total velocity. (And the number you should use for Vo is still just the up-down velocity that the object starts with.)
(published on 10/22/2007)
(published on 08/10/2009)
Driving force between two such objects will be gravitational pull: ma = -GmM/r2 First of all, let us simplify your problem slightly by assuming that the Sun is stationary. Because in reality, two objects would be accelerating towards their center of mass, as they will be pulled by equal magnitude forces. You can recover this effect by reduced mass concept.
d2r/dt2 = -GM/r2 Multiply by velocity and integrate w.r.t. time. Initial velocity is 0, initial distance R and let final position be r.
v dv/dt dt = -GM/r2 dv dt
v = sqrt( 2GM(R-r)/rR ) = -dr/dt This equation is separable, so integrate once more.
dr sqrt(rR) / sqrt (R-r) = -dt sqrt(2GM)
R3/2 ( arccos(sqrt(r/R)) - sqrt(r/R)(1-r2/R2) ) = t sqrt(2GM)
You are interested at the event where r(T)=0, so
R3/2 π/2 = T sqrt(2GM)
T = (π/2)sqrt(R3/2GM)
For your specific example, R = 1AU = 1.5 1011m, M = 2.1030kg, G = 6.7 10-11, which yields 6 106 seconds ~70 days.
And for another way of getting at the same result, less fundamental but maybe quicker, you can borrow Kepler's 3d law, that the orbital period is proportional to R3/2, where R is the long axis of the orbit ellipse. Falling into the sun is just half of an orbit whose ellipse just reaches the Sun, so with half the R of the Earth's yearly orbit. That whole other orbit would thus take 2-3/2 years so the falling time is 2-5/2 years, or ~65 days. Same answer, to our chosen accuracy. Mike W.
(published on 08/25/2015)