Gravitational Potential Energy and Total Energy
12.3 Gravitational Potential Energy and Total Energy
By the end of this section, you will be able to:
- Determine changes in gravitational potential energy over great distances
- Apply conservation of energy to determine escape velocity
- Determine whether astronomical bodies are gravitationally bound
Gravitational Potential Energy beyond Earth
Recall from Module 8 that we defined the change in potential energy as the negative work done by a conservative force:
When we calculated the change in gravitational potential energy, we assumed that the force of gravity was a constant force:
This was valid for objects moving near the surface of Earth because the force of gravity is very close to constant for small changes in height.
But now we need to consider how the gravitational potential energy is affected when objects with mass are moved farther apart or closer together, with distances that are large.
A student is studying the potential energy change of a 50 kg object raised 90 km above Earth’s surface. What will be the percentage error if she simply used the approximate relation ΔU = mgΔy?
A rock is dropped from a distance RE above the surface of the earth, and is observed to have kinetic energy K1 when it hits the ground. An identical rock is dropped from twice the height (2RE) above the earth’s surface and has kinetic energy K2 when it hits. RE is the radius of the earth.
What is K2 / K1?
|Compared to the earth, planet X has twice the mass and twice the radius. This means that compared to the amount of energy required to move an object from the earth’s surface to infinity, the amount of energy required to move that same object from planet X’s surface to infinity is|
|(a) four times as much.|
|(b) twice as much.|
|(c) the same.|
|(d) half as much.|
|(e) one-quarter as much.|
Bound and Unbound Systems
|A projectile is fired straight upward from the surface of an airless planet (radius R) with escape velocity. What is the projectile’s speed when it is a distance 4R from the planet’s center (3R from the surface)? (Ignore the gravity of the Sun and other astronomical bodies.)|
|(a) 1/2 vesc|
|(b) 1/4 vesc|
|(c) 1/9 vesc|
|(d) 1/3 vesc|
|(e) None of these is correct.|
|Does escape velocity depend on launch angle? |
(That is, if a projectile is given an initial speed vo, is it more likely to escape an (airless) planet, if fired straight up than if fired at an angle?)
The formula for energy conservation only involves v2, so angle does not matter.
If a planet has the same surface gravity as the earth (that is, the same value of g at the surface), what is its escape speed?
A planet orbiting a distant star has a radius 3.24 x 106 m. The escape speed for an object launched from this planet’s surface is 7.65 x 103 m/s. What is the acceleration due to gravity at the surface of the planet?