- #1

- 978

- 0

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?

Last edited: