Resource Lesson
Gravitational Potential Energy
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Recall that work is calculated with the formula
W = Fs cos θ
where θ is the angle between the direction of the force and the direction of the object's motion.
The work energy theorem states that the work done on an object equals the change in its kinetic energy,
W = ΔKE
.
When we examine an object dropped from rest in a state of freefall we notice that as gravity accelerates the object towards the surface of the earth, not only does the object's kinetic energy increases as its speed increases but its potential energy decreases as its height above the surface of the earth decreases. This is a general property of any conservative force: work done results in a subsequent decrease in the object's potential energy.
Work
_{done conservative force}
= -ΔU
Let's look at another example to highlight the inclusion of this "negative sign." Since we are working near the surface of the earth and our objects will not go beneath the surface, we are going to use the floor (the position A in the above diagram) as our zero position. Our initial discussions will employ the equation
PE
_{g}
= mgh
where
h = r - R
_{E}
. That is, our
h
will represent the difference between an object's initial distance from the center of the earth and the earth's surface. Later in the lesson we will see that this is a reasonable equation to calculate an object's gravitational potential energy when
r << R
_{E}
.
Note that the changes in an object's potential energy only depend on comparing its starting position and its ending position, not on whether it does or does not pass through various points in-between - that is, gravity is a
path-independent force
. The block's final change in potential energy is the same whether it follows the path with the intermediate stops B, C and D or whether it is directly taken from E to A. The height of the post is the same.
When
lowering
an object [(y
_{f}
- y
_{o}
) < 0], θ equals 0º.
θ equals 0º because the gravitational force (mg) and the direction of motion point in the same direction (cos 0º = 1)
the displacement h = (y
_{f}
- y
_{o}
) is negative - it is being lowered!
gravity would be doing this work since the natural direction of motion in a gravitational field is downward
Work
_{by gravity}
= mg(-h) cos 0º
= mg(-h)(+1)
= - mgh
The work done by gravity on the mass equaled the negative change in its potential energy, mgh.
Gravitational Potential Energy
For the remainder of the lesson we will use the variable
U
to represent potential energy as we derive a generalize equation for an object's potential energy at and arbitrary R > R
_{E}
.
Let's assume that our point mass begins at an extremely distant location infinitely far away and is being gravitationally attracted to a position closer to the earth's surface, R.
The vector nature of the gravitational force, F
_{G}
, is written as
where
is a unit vector pointing from M
_{1}
to M
_{2}
. Since gravity pulls M
_{2}
towards M
_{1}
, we use the negative to denote that the direction of the force is the reverse of
.
Recall from our discussion at the top of this page that the work done by gravity equals the negative change in the object's gravitational potential energy.
Since our force is not constant, we will have to use integration to calculate the work that will need to be done.
Putting together our two expressions for work we have
Simplifying the left side of our equation we get
Since our functional value depends on the ratio (1/R) it is natural to define the
potential energy (U
_{o}
) as being 0 at infinity
since the limit as R approaches infinity of (1/R) equals 0.
We now have our final expression for the potential energy of a satellite at any radius
R
from the center of a planet.
A graph of
U(R) vs R
is shown below.
As expected, all values of the potential energy are negative, approaching the value of zero as R approaches infinity. Because of this we say that the mass is trapped in an
"energy well"
- that is, it will have to be given additional energy from an external source if it is to escape gravity's attraction.
Refer to the following information for the next four questions.
A 10-kg satellite is orbiting 36 km above the surface of the earth in circular orbit.
What is the satelite's gravitational potential energy?
What is the satellite's kinetic energy?
What is the satellite's total energy?
What is the significance of the negative sign on your previous question?
Notice that the simplified expression for an object's potential energy near the surface of the earth that we used in the first part of our lesson, PE = mgh, can be derived from our generalized formula.
Recall that the gravitational field strength is given by
The potential energy at the surface of the earth is
Due to the almost linear nature of our graph near the planet's surface we can approximate the potential energy for small heights (h<<R
_{E}
) above the earth's surface as
The change in potential energy for a falling mass would be
Escape velocity
When a projectile is thrown upwards off the surface of the earth, the decrease in its kinetic energy equals the increase in its potential energy. At any given time, its total energy will equal
total energy = K + U
If we wanted an unpowered projectile to escape from the earth's surface, it would need to have a total energy equal to or greater than zero to get out of the earth's gravity well.
What is the escape velocity at the surface of the earth?
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