Lecture 38: Gravitational Potential • The force of between two objects depends only on the of the objects and the distance between them • If we take one of the objects to be fixed, the done by

gravity in moving the other object from distance r1 to r2 is: r2 = ( ) W12 — F r dr r1 and the work done to move it back to r2 is: r1 r2 = ( ) = − ( ) = − W21 — F r dr — F r dr W12 r2 r1

• So we see that the work done for a round trip from r1 to r2 and back is 0

Gravity is a conservative force, so we can define a associated with it • We can find the change in potential energy between r1 and r2 using the definition: r2 GMm ∆U = U (r )−U (r ) = −W = − dr 2 1 12 — r2 r1 »1ÿr2 »1 1 ÿ = GMm = GMm … − Ÿ …r ⁄Ÿ r r r1 2 1 ⁄ • As always, only differences in potential energy are physically important, and we are free to set U to 0 at any r we find convenient • The most common choice is to set r = ∞ . With that choice the potential energy at any other distance is: −GMm U (r) = r Escape Speed • The gravitational potential energy due to a single object looks like: 10 • If a particle has total energy > 0, the kinetic 5 energy is positive for all

0 0 values of r – Such a particle can -5 U(r) “escape” from the -10 gravitational attraction of the object -15

-20 • If a particle bergins a distance R from the object, the initial speed needed for escape is: 1 GMm 2GM E = mv2 − = 0; v = 2 R R

0 0.5 1 1.5 2 2.5 3 Example: Pioneer space probe • The Pioneer space probe was the first man-made object to leave the solar system – So it needed to escape from both the and the . Which requires the greatest initial velocity, if we start from the Earth’s surface? • Earth: 2GM 2 ⋅6.67 ×10−11 Nm2 /kg2 ⋅5.98×1024 kg v = E = × 6 rE 6.37 10 m = 11.2km/s

• Sun 2 ⋅6.67 ×10−11 Nm2 /kg2 ⋅1.99 ×1030 kg v = 1.50 ×1011 m = 42.1km/s Gravitational Potential Energy of Many- body Systems

• We can extend the two-body gravitational potential energy to an arbitrary number of particles by considering the work required to assemble the system “piece-by-piece” • Example: A three particle system • We want to find the gravitational potential energy of the following system: r 12 r23 m2 r m1 13 m3 • We begin with all three masses infinitely far from each other

• Then we move m1 to its final – We do no work here, since there is no gravitational force

• Next we bring in m2 at constant velocity, while m1 remains stationary – The work done is Gm m W = − 1 2 r12

– Negative since we are preventing m2 from accelerating toward m1

• Finally we bring in m3

• Now we have to fight against gravity from both m1 and m2, so the work done is: Gm m Gm m W = − 1 3 − 2 3 r13 r23 • This means that the total gravitational potential energy of the system is: Gm m Gm m Gm m U = − 1 2 − 1 3 − 2 3 r12 r13 r23 • In other words, it would take that much energy to move the masses infinitely far from each other – Note that the result does not depend on which order the masses are brought in from infinity, nor on the path taken while doing so • This type of potential energy is known as “binding energy”, and is a very useful concept in atomic and nuclear physics as well