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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

We studied gravitational potential energy in Potential Energy and Conservation of Energy , where the value of g remained constant. We now develop an expression that works over distances such that g is not constant. This is necessary to correctly calculate the energy needed to place satellites in orbit or to send them on missions in space.

Gravitational potential energy beyond earth

We defined work and potential energy in Work and Kinetic Energy and Potential Energy and Conservation of Energy . The usefulness of those definitions is the ease with which we can solve many problems using conservation of energy. Potential energy is particularly useful for forces that change with position, as the gravitational force does over large distances. In Potential Energy and Conservation of Energy , we showed that the change in gravitational potential energy near Earth’s surface is Δ U = m g ( y 2 y 1 ) . This works very well if g does not change significantly between y 1 and y 2 . We return to the definition of work and potential energy to derive an expression that is correct over larger distances.

Recall that work ( W ) is the integral of the dot product between force and distance. Essentially, it is the product of the component of a force along a displacement times that displacement. We define Δ U as the negative of the work done by the force we associate with the potential energy. For clarity, we derive an expression for moving a mass m from distance r 1 from the center of Earth to distance r 2 . However, the result can easily be generalized to any two objects changing their separation from one value to another.

Consider [link] , in which we take m from a distance r 1 from Earth’s center to a distance that is r 2 from the center. Gravity is a conservative force (its magnitude and direction are functions of location only), so we can take any path we wish, and the result for the calculation of work is the same. We take the path shown, as it greatly simplifies the integration. We first move radially outward from distance r 1 to distance r 2 , and then move along the arc of a circle until we reach the final position. During the radial portion, F is opposite to the direction we travel along d r , so E = K 1 + U 1 = K 2 + U 2 . Along the arc, F is perpendicular to d r , so F · d r = 0 . No work is done as we move along the arc. Using the expression for the gravitational force and noting the values for F · d r along the two segments of our path, we have

Δ U = r 1 r 2 F · d r = G M E m r 1 r 2 d r r 2 = G M E m ( 1 r 1 1 r 2 ) .

Since Δ U = U 2 U 1 , we can adopt a simple expression for U :

U = G M E m r .
An illustration of the earth and two larger concentric circles centered around it. The radius of the small circle is labeled r 1 with a black arrow and the radius of the larger circle is labeled r 2 with a black arrow. A red arrow extends from the end of the r 1 arrow to the larger circle, then forms an arc on the larger circle to the tip of the r 2 arrow. The red line is labeled Path of integration.
The work integral, which determines the change in potential energy, can be evaluated along the path shown in red.

Note two important items with this definition. First, U 0 as r . The potential energy is zero when the two masses are infinitely far apart. Only the difference in U is important, so the choice of U = 0 for r = is merely one of convenience. (Recall that in earlier gravity problems, you were free to take U = 0 at the top or bottom of a building, or anywhere.) Second, note that U becomes increasingly more negative as the masses get closer. That is consistent with what you learned about potential energy in Potential Energy and Conservation of Energy . As the two masses are separated, positive work must be done against the force of gravity, and hence, U increases (becomes less negative). All masses naturally fall together under the influence of gravity, falling from a higher to a lower potential energy.

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
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John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
David Reply
what is viscosity?
David
what is inorganic
emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
what's motion
Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
hello friend how are you
Muhammad Reply
fine, how about you?
Mohammed
hi
Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
Reofrir Reply
Practice Key Terms 2

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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