Where is Gravitational Potential Energy Zero?

AI Thread Summary
Gravitational potential energy (P.E) is relative and can be defined with respect to a chosen reference point. The discussion revolves around identifying where P.E is considered zero for a body in a planet's gravitational field, with options including the center of the planet, the surface, infinite distance, or at a distance equal to the Earth's radius. A common approach is to use the standard formula for gravitational potential energy, which inherently includes a reference point. Understanding this concept is crucial for solving related problems in gravitational physics. The key takeaway is that the definition of zero potential energy is dependent on the chosen reference point.
sphyics
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P.E of a body in the graitational field of planet is zero. The body must be at:

a) Centre of planet b) on the surface of planet c) Infinite distance d) At distance equal to radius of earth


how to think over the problem.
 
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sphyics said:
P.E of a body in the graitational field of planet is zero. The body must be at:

a) Centre of planet b) on the surface of planet c) Infinite distance d) At distance equal to radius of earth


how to think over the problem.

Potential energy is relative to some datum. You can define your datum wherever you want.
 
sphyics said:
how to think over the problem.
As berkeman says, you need a reference point. I suspect that they want you to use the standard formula for the gravitational PE between two bodies, which has a built-in reference point. What is that formula?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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