# Multivariable Calculus Problem

wittywonka
Science 12 May '10 03:58
1. wittywonka
Chocolate Expert
12 May '10 03:58
I am in AP Calculus in high school, but I have a question about a physics problem I made up that I assume is a multivariable calculus problem.

Force gravitational = GMm / (r^2)

Therefore,

Acceleration gravitational = GM / (r^2)

So, here's the problem I came up with. Let's say you take a basketball and travel out into space, approximately 1E10 meters from the center of the Sun, and you release the basketball from rest. Assume that the force of the Sun's gravity is the only force acting on the basketball at any time t. Also assume for the purposes of this problem that the product of G (gravitational constant) and M (mass of the Sun) is 1E20. Finally, assume that the ball is accelerating towards the sun in the negative x direction. Therefore, we know that at time 0, a(0) = -1 m/s^2, v(0) = 0 m/s, and s(0) = 1E10 m.

How would I go about deriving an equation to calculate the velocity of the basketball at any time t? Both time and distance/radius are variables, so again I assume this is a multivariable problem.

Any explanations in addition to any answers would be greatly appreciated.
2. 12 May '10 07:00
Since the velocity at time = 0 is 0, and the acceleration is in the negative x direction at this point, it follows that the trajectory of the object is parallel to the x-axis. We can therefore rewrite F=GMm/x^2.

See if you can come up with a differential equation which has time and x. I suspect this would be the correct way to start this problem.
3. 12 May '10 07:111 edit
4. 12 May '10 09:48
Using Newton's second law, you can write F = ma, so a = GM/x² like you say. Since a = dx²/dt², you obtain a differential equation that you can then solve using the appropiate boundary conditions:

dx²/dt² = GM/x²

Once you obtain x, you can take the derivative to obtain v. I don't know the solution to this equation immediately, however.
5. 12 May '10 11:17
It turns out the solution is pretty complicated, and I can only get Mathematica to give me a fairly lengthy, implicit equation for x. On a side note, you don't need a boundary condition for a(0) - you need as many boundary conditions as you have derivatives in your differential equation.
6. 12 May '10 15:34
You could try an energy-based approach instead, since it's easy enough to calculate the potential energy as a function of x. That gives you a first-order equation to solve for x(t) which might be easier.
7. joe shmo
Strange Egg
12 May '10 20:40
Originally posted by KazetNagorra
Using Newton's second law, you can write F = ma, so a = GM/x² like you say. Since a = dx²/dt², you obtain a differential equation that you can then solve using the appropiate boundary conditions:

dx²/dt² = GM/x²

Once you obtain x, you can take the derivative to obtain v. I don't know the solution to this equation immediately, however.
is this a non-linear diff eq?
8. 12 May '10 21:17
Originally posted by joe shmo
is this a non-linear diff eq?
Yes.
9. 13 May '10 04:49
What seperates Humanity from the Animals is our ability to calculate complex derivatives of compound triganomic functions using implicit differentiation!
10. 13 May '10 07:31
Despite the nonlinearity of the equation, since its separable shouldn't the solution be fairly straightforward? Or am I missing something here?
11. 13 May '10 09:41
Originally posted by amolv06
Despite the nonlinearity of the equation, since its separable shouldn't the solution be fairly straightforward? Or am I missing something here?
Hadn't thought of that, yes, that should solve the equation quite easily.

So you have dx²/dt² = GM/x²
x² dx² = GM dt²
(1/3) x³ dx + C = GM t dt
(1/12) x^4 + Cx + D = GM (1/2) t²
12. 13 May '10 09:50
Originally posted by KazetNagorra
Hadn't thought of that, yes, that should solve the equation quite easily.

So you have dx²/dt² = GM/x²
x² dx² = GM dt²
(1/3) x³ dx + C = GM t dt
(1/12) x^4 + Cx + D = GM (1/2) t²
'Fraid that doesn't work 🙂. You can't do that with a second derivative.
13. 13 May '10 09:58
Originally posted by mtthw
'Fraid that doesn't work 🙂. You can't do that with a second derivative.
Well, that would explain why Mathematica's solution is a bit more messy!
14. Palynka
Upward Spiral
13 May '10 13:21
Originally posted by KazetNagorra
Well, that would explain why Mathematica's solution is a bit more messy!
What is Mathematica's solution?
15. Agerg
The 'edit'or
13 May '10 14:181 edit
Originally posted by Palynka
What is Mathematica's solution?
I don't use mathematica but the tex output for Maxima's solution is as follows (rather ugly):

$$\frac{K\,M\,log\left( \frac{\sqrt{\frac{\left( \%k1\,x-1\right) \,K\,M}{x}}-\sqrt{\%k1\,K\,M}}{\sqrt{\frac{\left( \%k1\,x-1\right) \,K\,M}{x}}+\sqrt{\%k1\,K\,M}}\right) -2\,x\,\sqrt{\%k1\,K\,M}\,\sqrt{\frac{\left( \%k1\,x-1\right) \,K\,M}{x}}}{2\,\sqrt{2}\,\%k1\,K\,M\,\sqrt{\%k1\,K\,M}}=t+\%k2,-\frac{K\,M\,log\left( \frac{\sqrt{\frac{\left( \%k1\,x-1\right) \,K\,M}{x}}-\sqrt{\%k1\,K\,M}}{\sqrt{\frac{\left( \%k1\,x-1\right) \,K\,M}{x}}+\sqrt{\%k1\,K\,M}}\right) -2\,x\,\sqrt{\%k1\,K\,M}\,\sqrt{\frac{\left( \%k1\,x-1\right) \,K\,M}{x}}}{2\,\sqrt{2}\,\%k1\,K\,M\,\sqrt{\%k1\,K\,M}}=t+\%k2]$$
where \%k1 and \%k2 are constants of integration