# Rate of change of acceleration due to gravity **URL:** \ **Category:** Factual Questions\ **Created:** [April 24, 2008, 10:14pm UTC](https://boards.straightdope.com/t/rate-of-change-of-acceleration-due-to-gravity/446748 "2008-04-24T22:14:03Z")\ **Posts on this page:** 15\ **Page:** 1
**Author:** ![sweeteviljesus](https://avatars.discourse-cdn.com/v4/letter/s/898d66/32.png) [@sweeteviljesus](https://boards.straightdope.com/u/sweeteviljesus)\ **Post date:** [April 24, 2008, 10:14pm UTC](https://boards.straightdope.com/t/rate-of-change-of-acceleration-due-to-gravity/446748/1 "2008-04-24T22:14:03Z")
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 ---
**Author:** ![mks57](https://avatars.discourse-cdn.com/v4/letter/m/b9bd4f/32.png) [@mks57](https://boards.straightdope.com/u/mks57)\ **Post date:** [April 24, 2008, 10:35pm UTC](https://boards.straightdope.com/t/rate-of-change-of-acceleration-due-to-gravity/446748/2 "2008-04-24T22:35:47Z")
[Jerk](http://en.wikipedia.org/wiki/Jerk)? ---
**Author:** ![sweeteviljesus](https://avatars.discourse-cdn.com/v4/letter/s/898d66/32.png) [@sweeteviljesus](https://boards.straightdope.com/u/sweeteviljesus)\ **Post date:** [April 24, 2008, 10:45pm UTC](https://boards.straightdope.com/t/rate-of-change-of-acceleration-due-to-gravity/446748/3 "2008-04-24T22:45:17Z")
Well, I was looking for the solved version of the equation, i.e. with no derivative terms in it. Thanks, Rob ---
**Author:** ![sweeteviljesus](https://avatars.discourse-cdn.com/v4/letter/s/898d66/32.png) [@sweeteviljesus](https://boards.straightdope.com/u/sweeteviljesus)\ **Post date:** [April 24, 2008, 10:46pm UTC](https://boards.straightdope.com/t/rate-of-change-of-acceleration-due-to-gravity/446748/4 "2008-04-24T22:46:56Z")
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 ---
**Author:** ![TJdude825](https://avatars.discourse-cdn.com/v4/letter/t/dec6dc/32.png) [@TJdude825](https://boards.straightdope.com/u/TJdude825)\ **Post date:** [April 24, 2008, 11:14pm UTC](https://boards.straightdope.com/t/rate-of-change-of-acceleration-due-to-gravity/446748/5 "2008-04-24T23:14:09Z")
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. ---
**Author:** ![sweeteviljesus](https://avatars.discourse-cdn.com/v4/letter/s/898d66/32.png) [@sweeteviljesus](https://boards.straightdope.com/u/sweeteviljesus)\ **Post date:** [April 24, 2008, 11:19pm UTC](https://boards.straightdope.com/t/rate-of-change-of-acceleration-due-to-gravity/446748/6 "2008-04-24T23:19:56Z")
Doesn’t that take us back to where we started? Thanks, Rob ---
**Author:** ![si\_blakely](https://avatars.discourse-cdn.com/v4/letter/s/d9b06d/32.png) [@si\_blakely](https://boards.straightdope.com/u/si_blakely)\ **Post date:** [April 25, 2008, 11:33am UTC](https://boards.straightdope.com/t/rate-of-change-of-acceleration-due-to-gravity/446748/7 "2008-04-25T11:33:43Z")
[QUOTE=sweeteviljesus] Specifically, my dad wanted to know what the rate of acceleration of a body toward the sun would be, ignoring other massive bodies. [/QUOTE] 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](http://www.aerospaceweb.org/question/astronomy/q0227.shtml).) 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 ---
**Author:** ![SenorBeef](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/senorbeef/32/2220_2.png) [@SenorBeef](https://boards.straightdope.com/u/SenorBeef)\ **Post date:** [April 25, 2008, 11:43am UTC](https://boards.straightdope.com/t/rate-of-change-of-acceleration-due-to-gravity/446748/8 "2008-04-25T11:43:44Z")
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? ---
**Author:** ![SenorBeef](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/senorbeef/32/2220_2.png) [@SenorBeef](https://boards.straightdope.com/u/SenorBeef)\ **Post date:** [April 25, 2008, 11:58am UTC](https://boards.straightdope.com/t/rate-of-change-of-acceleration-due-to-gravity/446748/9 "2008-04-25T11:58:18Z")
Meant to post that to a new thread. Can someone report that post for deletion? ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [April 25, 2008, 12:14pm UTC](https://boards.straightdope.com/t/rate-of-change-of-acceleration-due-to-gravity/446748/10 "2008-04-25T12:14:21Z")
[QUOTE=si\_blakely] 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](http://www.aerospaceweb.org/question/astronomy/q0227.shtml).) [/QUOTE] 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. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [April 25, 2008, 12:17pm UTC](https://boards.straightdope.com/t/rate-of-change-of-acceleration-due-to-gravity/446748/11 "2008-04-25T12:17:33Z")
[QUOTE=SenorBeef] Meant to post that to a new thread. Can someone report that post for deletion? [/QUOTE] Done. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [April 25, 2008, 12:33pm UTC](https://boards.straightdope.com/t/rate-of-change-of-acceleration-due-to-gravity/446748/12 "2008-04-25T12:33:21Z")
[QUOTE=zut] If I’m doing this right (some one check my work), the function should be… [/quote] 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. ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [April 25, 2008, 1:52pm UTC](https://boards.straightdope.com/t/rate-of-change-of-acceleration-due-to-gravity/446748/13 "2008-04-25T13:52:58Z")
[QUOTE=si\_blakely] 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). [/QUOTE] 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. ---
**Author:** ![si\_blakely](https://avatars.discourse-cdn.com/v4/letter/s/d9b06d/32.png) [@si\_blakely](https://boards.straightdope.com/u/si_blakely)\ **Post date:** [April 28, 2008, 8:32am UTC](https://boards.straightdope.com/t/rate-of-change-of-acceleration-due-to-gravity/446748/14 "2008-04-28T08:32:54Z")
[QUOTE=Q.E.D.] 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. [/QUOTE] You’re right :smack: - too much terrestrial physics. Sigh Si ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [April 28, 2008, 1:49pm UTC](https://boards.straightdope.com/t/rate-of-change-of-acceleration-due-to-gravity/446748/15 "2008-04-28T13:49:20Z")
At least you’re well grounded in the subject. 😉