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.
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.
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.
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Originally posted by KazetNagorrais this a non-linear diff eq?
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.
Originally posted by amolv06Hadn't thought of that, yes, that should solve the equation quite easily.
Despite the nonlinearity of the equation, since its separable shouldn't the solution be fairly straightforward? Or am I missing something here?
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
Originally posted by KazetNagorra'Fraid that doesn't work 🙂. You can't do that with a second derivative.
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
Originally posted by PalynkaI don't use mathematica but the tex output for Maxima's solution is as follows (rather ugly):
What is Mathematica's solution?
$$\frac{K\,M\,log\left( \frac{\sqrt{\frac{\left( \%k1\,x-1\right) \,K\,M}{x}}-\sqrt{\%k1\,K\,M}}{\sqrt{\frac{\left( \%k1\,x-1\right) \,K\,M}{x}}+\sqrt{\%k1\,K\,M}}\right) -2\,x\,\sqrt{\%k1\,K\,M}\,\sqrt{\frac{\left( \%k1\,x-1\right) \,K\,M}{x}}}{2\,\sqrt{2}\,\%k1\,K\,M\,\sqrt{\%k1\,K\,M}}=t+\%k2,-\frac{K\,M\,log\left( \frac{\sqrt{\frac{\left( \%k1\,x-1\right) \,K\,M}{x}}-\sqrt{\%k1\,K\,M}}{\sqrt{\frac{\left( \%k1\,x-1\right) \,K\,M}{x}}+\sqrt{\%k1\,K\,M}}\right) -2\,x\,\sqrt{\%k1\,K\,M}\,\sqrt{\frac{\left( \%k1\,x-1\right) \,K\,M}{x}}}{2\,\sqrt{2}\,\%k1\,K\,M\,\sqrt{\%k1\,K\,M}}=t+\%k2]$$
where \%k1 and \%k2 are constants of integration