Aim: How do we explain energy gravitational potential energy?

Gravitational Potential Energy The gravitational potential energy of a system can be found by calculate the work done by the force of gravity over a path Ug=∫Fgdr Ug =-GMm/r

Two points about gravitational potential energy What happened to our equation for gravitational potential energy, Ug = mgh equation? This equation only applies to scenarios where gravity can be expressed as Fg=mg which is near the surface of the Earth Why is the gravitational potential energy equation, Ug = -Gm1m2/r negative? It is a bounding energy

Graph of the Potential Energy Function Between Two Masses What happens to the gravitational potential energy between a mass and a planet as the distance approaches infinity? It approaches 0

Conservation of Energy of a System Energy is conserved. E = K + Ug =1/2mv2 – GMm/r = constant

Problem 1 A satellite of the Earth had a mass of 100 kg and its distance is 8.36 x 106 m from the center of the Earth (2 x 106 m from the surface). What is the gravitational potential energy of the satellite-Earth system? Ug=-GMm/r = -(6.67 x 10-11 )(5.98 x 1024 )(100)/(2 x 106 ) = -4.8 x 109 J b)What is the magnitude of the force on the satellite? Fg =GMm/r2 =(6.67 x 10-11)(5.98 x 1024)(100)/(2x106)2 =570 N a)-4.8 x10^9 J b) 570 N

Thought Question 1 From an inertial frame in space, we watch two identical uniform spheres fall toward one another owing to their mutual gravitational attraction. Approximate their initial speed as zero and take the initial gravitational potential energy of the two sphere system as Ui. When the separation between the two spheres is half the initial separation, what is the kinetic energy of each sphere? Since U=-GMm/r when the separation is halved, the potential energy will double. Thus we apply energy conservation and find: -Ui=-2Ui + KE1+KE2 Ui=KE + KE Ui=2KE so KE =Ui/2

Thought Question 2 The figure shows six paths by which a rocket orbiting a moon might move from point a to point b. Rank the paths according to (a) the corresponding change in gravitational potential energy of the rocket-moon system and (b) the net work done on the rocket by the gravitational force from the moon, greatest first. They are all the same

Thought Question 2

Thought Question 3 A particle of mass m(not shown) is to be moved from an infinite distance to one of the three possible locations a,b,and c. Two other particles of masses m and 2m, are fixed in place. Ranks the three possible locations according to the work done by the net gravitational force on the moving particle due to the fixed particles, greatest first. B>C>A

Thought Question 3

Circular Orbit Express the kinetic energy of the orbiting mass. KE=1/2mv2=1/2m(GM/r)=1/2GMm/r Express the gravitational potential energy of the orbiting mass. U=-GMm/r Express the total energy of the orbiting mass. E=KE+U=-GMm/2r

Escape Speed 1/2mv2 –GMEm/RE = -GMEm/rmax Taking rmax to be infinity, 1/2mv2 = GMEm/RE vesc = √(2GME/R) In general, Thus, vesc = √(2GM/R)

Thought Question 4 If you were a planetary prospector and discovered gold on an asteroid, it probably would not be a good idea to jump up and down in excitement over your find. Why? If you jump up at too fast a speed, you may exceed the escape velocity

Problem 2 The escape speed from the surface of the Earth is 11.2 km/s. Estimate the escape speed for a spacecraft from the surface of the Moon. The moon has a mass of 1/81 that of Earth and a radius ¼ that of the Earth. V=√[2G(1/81 M)/(1/4 R)]= 2/9 √[2GM/R] = 2/9 (11.2 km/s) = 2.49 km/s 2.49 km/s