If the gravitational constant g decreases and approaches 0, why does PE inc?

[yellowcocopuffs]

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EK physics problem 67 on 30 min exams

It says that as the g decreases and approaches 0, the grav PE of rocket continually increases. However, since PE=mgh and g approaches 0, wouldn't PE approach 0 also?

 

[Rabolisk]

mgh is typically used only for when g is constant, i.e. over a short range of distance, usually near the surface of the Earth where it is conveniently about 9.8m/s^2. You have to look at U = -Gm1m2/r. Let's say that m1 is the mass of the earth and m2 is the mass of the rocket. The common g that you encounter in mgh is derived from -Gm1/r using the radius of the earth as r. However, when you go farther out from the earth, distance increases, decreasing the value of g, or the acceleration due to gravity. Also, U, the potential energy increases as you increase r. Don't forget the negative sign.

Intuitively, just know that potential energy always increases as the distance from the center of the Earth increases. Three ways to increase potential energy are to increase mass of the object, mass of the planet/star, and increase the distance between them.

 

[sangria1986]

EK physics problem 67 on 30 min exams

It says that as the g decreases and approaches 0, the grav PE of rocket continually increases. However, since PE=mgh and g approaches 0, wouldn't PE approach 0 also?

for a rocket mgh no longer applies. remember, the concept of mg is actually from Gravity force =m1*m2/r^2 (if you plug in the values of radius of earth r and mass of earth you will find that the value is 9.8 m/s^2 * some value mass...

For a rocket, the radius is great enough that mgh cannot be used and the above equation has to be.

This equation represents the FORCE of gravity....however, they are talking about potential energy. We know that potential energy =- Work. The work done by gravity force is above equation *d or r in this case. Thus work done by gravity = Gmm/r. however, potential energy = NEGATIVE of work. or U=-Gmm/r.

Since r is in the denominator and since the overall equation is negative, as R increases the value -Gmm/r goes to ZERO....thus as radius increases the potential energy is actually increasing-from a negative value to zero.

You might ask what is positive energy then? In this case...there is no such thing as positive gravitational potential energy because zero represents the energy to remove an object from earths gravity at infinite distance.