If I start with F=ma and F=GmM/r^2, I get a=GM/r^2. To get da/dt, what do I need to substitute in for r? r[0] + v[0]t + 1/2at^2? How do you deal with the a on both sides of the equation in that case? This is not a homework problem, BTW. My dad asked me this one time and I was curious what the third derivative of position was.
Thanks for your help,
Rob
Jerk?
Well, I was looking for the solved version of the equation, i.e. with no derivative terms in it.
Thanks,
Rob
Specifically, my dad wanted to know what the rate of acceleration of a body toward the sun would be, ignoring other massive bodies.
Thanks,
Rob
a=GM/r^2 is a 2nd order differential equation. It doesn’t look like one, maybe, but a is the 2nd derivative of r. So:
d[sup]2[/sup]r / dt[sup]2[/sup] = GM/r[sup]2[/sup]
I THINK you would solve this by multiplying both sides by r[sup]2[/sup], then integrating both sides twice. This would give you (after a lot of ugly things like integration by parts, probably) r, as a function of t. Then you can find a, as a function of t, by simply differentiating that answer twice, or writing a(t) = GM/[r(t)][sup]2[/sup]. Either way, differentiate a(t) to get da/dt.
Doesn’t that take us back to where we started?
Thanks,
Rob
You are asking 2 questions here inadvertently, and you have answered one of them. a=GM/r[sup]2[/sup]. This is a constant. The acceleration of a body towards the sun is a constant (just like a body falling towards earth has a constant acceleration). For the sun this is 274.13 m/s[sup]2[/sup] (from here.)
The third derivative of position is sometimes called jerk, the rate of change of acceleration, as noted. In the case of gravitational forces, this is zero. For jerk to be a nonzero derivative, position needs to defined by at least a cubic function wrt t.
Si
I was watching a show that partly documented some of the problems with extended periods of weightlessness. They mentioned that there was a regulatory system that changed the bloodflow based on whether the body was in a horizontal/supine position or standing. I don’t have more context than that.
Is that true? If so, what is the mechanism responsible for this called and how does it work?
Meant to post that to a new thread. Can someone report that post for deletion?
You’re misinterpreting hte question. What you say is (approximately) true for a limited range of motion near the surface of the sun. The OP is asking, “what happens if an object falls toward the sun from a very long distance away”? In that case, the acceleration changes (because r changes), and the jerk is non-zero.
If I’m doing this right (some one check my work), the function should be
r = (GM)[sup]1/3[/sup]t[sup]2/3[/sup]
Then
dr/dt (velocity) = (GM)[sup]1/3[/sup]t[sup]-1/3[/sup]
d[sup]2[/sup]r/dt[sup]2[/sup] (accel) = (GM)[sup]1/3[/sup]t[sup]-4/3[/sup]
d[sup]2[/sup]r/dt[sup]2[/sup] (jerk) = (GM)[sup]1/3[/sup]t[sup]-7/3[/sup]
Note also that, referring to TJdude825’s post,
r[sup]2[/sup] = (GM)[sup]2/3[/sup]t[sup]4/3[/sup]
so GM/r[sup]2[/sup] = (GM)[sup]1/3[/sup]t[sup]-4/3[/sup], which is indeed the acceleration.
Done.
Gah! I did screw it up.
r = -(9/2GM)[sup]1/3[/sup]t[sup]2/3[/sup]
Then
dr/dt (velocity) = -2/3(9/2GM)[sup]1/3[/sup]t[sup]-1/3[/sup]
d[sup]2[/sup]r/dt[sup]2[/sup] (accel) = 2/9(9/2GM)[sup]1/3[/sup]t[sup]-4/3[/sup]
d[sup]2[/sup]r/dt[sup]2[/sup] (jerk) = 8/27(9/2GM)[sup]1/3[/sup]t[sup]-7/3[/sup]
Note also that, referring to TJdude825’s post,
r[sup]2[/sup] = (9/2GM)[sup]2/3[/sup]t[sup]4/3[/sup]
so GM/r[sup]2[/sup] = (GM)[sup]1/3/sup[sup]-2/3[/sup]t[sup]-4/3[/sup] = 2/9(9/2GM)[sup]1/3[/sup]t[sup]-4/3[/sup], which is indeed the acceleration.
I think.
No, it isn’t. G is a constant. For a given body, M is also a constant. But, as r changes, so does the acceleration due to gravity. The site you linked to deals with surface gravity. We don’t normally bother to include this in calculation dealing with ordinary freefall conditions, since the change in r is very small over the distances involved because r includes the radius of the Earth.
You’re right :smack: - too much terrestrial physics.
Sigh
Si
At least you’re well grounded in the subject. 