- #1
kmarinas86
- 979
- 1
For the equation:
[itex]U=\frac{-GMm}{h}[/itex]
Where [itex]h[/itex] is the distance between the center of masses [itex]M[/itex] and [itex]m[/itex].
In Calculus, they teach you derivatives.
The derivative of [itex]U[/itex] with respect to [itex]h[/itex] is:
[itex]dU=d\left(\frac{-GMm}{h}\right)[/itex]
[itex]dU=\frac{GMm}{h^2}[/itex]
Which is the gravitational force.
Were I to apply this knowledge to the pioneer anomaly, I would deduce that the gravitational potential energy would be equal to the integral of the force with respect to [itex]h[/itex]:
[itex]g_{pioneer}=8.74*10^{-10}\frac{m}{s^2}[/itex]
[itex]dU=\frac{GMm}{h^2}+mg_{pioneer}[/itex]
[itex]dU=d\left(\frac{-GMm}{h}+mg_{pioneer}h\right)[/itex]
[itex]U=\frac{-GMm}{h}+mg_{pioneer}h[/itex]
Are my premises true?
[itex]U=\frac{-GMm}{h}[/itex]
Where [itex]h[/itex] is the distance between the center of masses [itex]M[/itex] and [itex]m[/itex].
In Calculus, they teach you derivatives.
The derivative of [itex]U[/itex] with respect to [itex]h[/itex] is:
[itex]dU=d\left(\frac{-GMm}{h}\right)[/itex]
[itex]dU=\frac{GMm}{h^2}[/itex]
Which is the gravitational force.
Were I to apply this knowledge to the pioneer anomaly, I would deduce that the gravitational potential energy would be equal to the integral of the force with respect to [itex]h[/itex]:
[itex]g_{pioneer}=8.74*10^{-10}\frac{m}{s^2}[/itex]
[itex]dU=\frac{GMm}{h^2}+mg_{pioneer}[/itex]
[itex]dU=d\left(\frac{-GMm}{h}+mg_{pioneer}h\right)[/itex]
[itex]U=\frac{-GMm}{h}+mg_{pioneer}h[/itex]
Are my premises true?
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