Gravitational Force, the derivative of Gravitational Potential Energy?by kmarinas86 Tags: derivative, energy, force, gravitational, potential 

#1
Mar2706, 05:56 PM

P: 1,011

For the equation:
[latex]U=\frac{GMm}{h}[/latex] Where [latex]h[/latex] is the distance between the center of masses [latex]M[/latex] and [latex]m[/latex]. In Calculus, they teach you derivatives. The derivative of [latex]U[/latex] with respect to [latex]h[/latex] is: [latex]dU=d\left(\frac{GMm}{h}\right)[/latex] [latex]dU=\frac{GMm}{h^2}[/latex] 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 [latex]h[/latex]: [latex]g_{pioneer}=8.74*10^{10}\frac{m}{s^2}[/latex] [latex]dU=\frac{GMm}{h^2}+mg_{pioneer}[/latex] [latex]dU=d\left(\frac{GMm}{h}+mg_{pioneer}h\right)[/latex] [latex]U=\frac{GMm}{h}+mg_{pioneer}h[/latex] Are my premises true? 



#2
Mar2706, 06:45 PM

Mentor
P: 40,889

It's not clear what you are trying to do. For one thing, the pioneer anomaly refers to an unexplained residual acceleration after all known forces (like gravity from known masses) have been accounted for.




#3
Mar2706, 07:35 PM

P: 1,011

Main point:
Is the Gravitational Force the derivative of Gravitational Potential Energy? A motive: In case if this is correct, this would be provide information of its truth. Secondary (following) point: Is the main point still true if the pioneer anomalous acceleration is added into the acceleration due to gravity? Note that if the pioneer anomaly exists, it invalidates previous estimates of the sun's mass. A motive: In case if this is false, this would be provide information of its falsity. 



#4
Mar2706, 07:48 PM

Mentor
P: 40,889

Gravitational Force, the derivative of Gravitational Potential Energy?Are you trying to model the additional solar mass needed to account for the anomalous acceleration? If so, no need to work so hard. If the sun had an extra mass [itex]\Delta M[/itex], the additional acceleration would be: [tex]\Delta g = \frac{G \Delta M}{h^2}[/tex] where h is the distance from the sun's center to the pioneer. 


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