Does gravitational potential energy equation account for Earth's varying gravity?

[Miraj Kayastha]

When a rocket is launched into space, its GPE increases and its KE decreases.

If we equate the change in both energies we can find out the velocity of the rocket at a certain height.
Δ GPE = Δ KE

Does the equation already take the opposing gravitational force of Earth into account?

 

[Nugatory]

When a rocket is launched into space, its GPE increases and its KE decreases.

If we equate the change in both energies we can find out the velocity of the rocket at a certain height.
Δ GPE = Δ KE

Does the equation already take the opposing gravitational force of Earth into account?

Yes. The GPE is represents the work done against the gravitational force of the earth. If it weren't for that force, the potential energy wouldn't increase and the kinetic energy wouldn't decrease.

 

[Miraj Kayastha]

Isnt GPE the work done in bringing a mass from infinity to a point inside a particular gravitational field?

 

[Nugatory]

Isnt GPE the work done in bringing a mass from infinity to a point inside a particular gravitational field?

No, it's defined as the work done in bringing a mass from one point to another in a particular gravitational field, plus the PE at the initial point. The value of the PE that we assign to that initial point is completely arbitrary.

You are thinking of the very useful and widely used convention in which we choose to assign zero to the potential energy at infinity. Then the PE at any other point is, as you say, the (negative of) the work done on bringing a mass in from infinity. However, we could just as reasonably choose to assign zero potential to a point at the surface of the earth, and this choice makes the calculations easier if you're working with a rocket launched from the surface of the earth.

 

[Miraj Kayastha]

Does changing the position of zero position change the formula of GPE?

 

[jtbell]

Yes. If we define GPE = 0 at r = ∞, then
$$GPE = -\frac{GMm}{r}$$
where M is the mass of the earth, m is the mass of the object in question, and r is the distance from the center of the earth. (This applies only above the surface of the earth)

If we define GPE = 0 at r = R, where R is the radius of the earth, then
$$GPE = GMm \left( \frac{1}{R} - \frac{1}{r} \right)$$

You should be able to check for yourself that both formulas give the same result for ΔGPE between two different values of r.

 

[Miraj Kayastha]

In this case of rocket launch what force is doing work?

 

[Nugatory]

In this case of rocket launch what force is doing work?

Gravity is doing work on the rocket, and the thrust of the rocket motor is doing work on the rocket.

 

[Miraj Kayastha]

So can the potential energy of the rocket can be calculated by the formula

(Thrust - Gravitational force) * distance

In my book the gain in GPE of the rocket is calculated by -GMm/r which basically came from integrating Gravitational force* distance

Is there any difference?

I am really confused in the case of rocket launching

 

[jbriggs444]

Potential energy applies when the work it takes to reach a certain position against a particular force does not depend on the path you take to get there. Gravity has this property. Gravitational force is a "conservative field". So it makes sense to talk about the potential energy of gravity. Thrust does not have this property. So it does not make sense to talk about the potential energy of thrust.

 

[jtbell]

So can the potential energy of the rocket can be calculated by the formula

(Thrust - Gravitational force) * distance

No, because the rocket thrust is not a conservative force. You can define potential energy only for conservative forces such as gravity.