Download PES 1110 Fall 2013, Spendier Lecture 37/Page 1 Today

PES 1110 Fall 2013, Spendier
Lecture 37/Page 1
Today:
- Gravitational Potential Energy (13.6)
Recap last lecture
Acceleration due to Gravity
Newton proposed that the acceleration due to gravity is given by
Gm planet
aG 
(towards the center of the planet)
r2
ag = acceleration
G = Gravitational constant = 6.67 x 10-11 N m2 kg-2 (m3 kg-1 s-2)
mplanet = mass of planet
r = distance from center of planet
If you put in the mass of the Earth and the radius of the Earth in the above equation, we
obtain our g = 9.8 m/s2, we used since the 2nd day of class. Originally when capital G was
first measured, we knew what little g was already from experiments. We also knew the
radius of the earth so the equitation above was first used to calculate the mass of the
earth. This equation works on any planet!
Newton’s Law of Gravitation
Every object with mass exerts a gravitational force on every other object with mass.
For an object with mass M1 located r away from mass M2 (before we called this mass
mplanet)
FG  M1 aG
GM 1 M 2
FG 
(call this an inverse square law)
r2
r = separation distance, center-to-center for spherical objects
For a system of particles: Principle of Superposition
For n interacting particle, the gravitational force on particle 1 is:
n 





F1, net  F2on1  F3on1  F4on1  ...  Fn on 1   Fi on 1
i 2
PES 1110 Fall 2013, Spendier
Lecture 37/Page 2
Gravitational Potential Energy
Now also have to modify our equation for gravitational potential energy.
U = Mgy
Only works for problems when you are negligibly high above the surface, for distances
y ≪ RP (radius of planet), i.e. for distances smaller than the radius of the planet when
measured from the center of the planet. For distances larger compared to the radius, we
have to start over.
Here we have a planet of mass M1 and a satellite of mass M2 very far away compared to
the radius measured from the center of the planet. We now have to start over and
calculate the work done by gravity as the distance between mass M1 and M2 changes from
radius r1 to r2.
We can do this now, since all of you have enough calculus done. We need calculus since
the force is not constant any more!
FG 
GM 1 M 2
r2
it is an inverse square law!
This is definitely not a constant force. If we double the distance, the force decreases by a
factor of 4.
 
Work for constant force: W  F  d  Fd cos q
When force is not constant, we need to do integration: the work done by gravity
as the distance between mass M1 and M2 changes from radius r1 to r2 is
r2
 
Wg   Fg  dr
r1
(dot product between the gravitational force Fg and some dr, change in radius)
PES 1110 Fall 2013, Spendier
Lecture 37/Page 3
We now sweep some vector calculus under the rug here, because we do not need it.
In what direction is the gravitational force on mass M2. It is towards the center of mass
M1. Since the force is towards the center of M1 all we need to worry about is the change
in size of radius r. How does r get bigger? By going away from the planet, we increase


the center-to-center distance between the two masses. So the angle between Fg and dr is
180º.
r2
r2
  r2
GM 1M 2
Wg   Fg  dr   Fg dr cos(180)  
dr
r2
r
r
r
1
1
1
You know that when you take a dot product you take the cosine between the two vectors.
r2
Wg  
r1
r
r
2
 1  2 GM 1M 2 GM1M 2
GM1M 2
1
  
dr


GM
M
dr


GM
M

1
2
1
2
2
 r 
r2
r
r
r1
2
r
r
1
1
Since we know that
Wg  U g  U g 2  U g1
we can write by inspection
GM1 M 2
U g 2 
r2
And
GM 1 M 2
U g1  
r1
Therefore that gravitational potential energy is given by:
Ug  
GM 1M 2
r
[J]
You will not need to know this proof. But you should know that for each force there is a
corresponding work/energy. You also need to know that gravitational force goes as 1/r2
and gravitation potential energy goes as 1/r. You also need to know that there is a
negative sign.
PES 1110 Fall 2013, Spendier
Lecture 37/Page 4
The Negative Sign
Ug  
GM 1M 2
r
[J]
The reason for the negative sign is that it comes with a built on zero. Where is
gravitational potential energy zero? As r  , U g  0 (limit of the above function)
Plot of U g ( r ) as a function of r
When M1 and M2 are infinitely far apart Ug = 0. When objects move in the direction of
the force acting on them, their potential energy decreases  as M2 gets closer to M1, its
potential energy decreases from zero to a negative amount.
Example: When I drop an eraser from a given height and it hits the ground the potential
energy decreased. Just like the eraser, mass M2 wants to get closer to mass M1, because
the gravitational force is an attractive force. So we see that the gravitational potential
energy increases with increasing r, just like we saw when we used U = mgh on the
surface of the earth.
PES 1110 Fall 2013, Spendier
Lecture 37/Page 5
Example 1:
What is the gravitational potential energy between the Earth and the moon?
Now we can return to doing problems where one from of energy is converted to another.
Mainly, we will have gravitational potential energy being converted to kinetic energy.
Conservation of Energy
When gravity is the only force doing work on a rocket with mass M near a planet, MP :
Ki + Ugi = Kf + Ugf
1
GM P M 1
GM P M
Mvi2 
 Mv 2f 
2
ri
2
rf
vi = initial velocity of the rocket
ri = initial separation distance for the center of the rocket to center of the planet
vf = final velocity of the rocket
rf = final separation
One of the many things we can use this for is to calculate something called the escape
speed
Escape speed:
The initial speed needed by a rocket in order to barely escape from a planet’s gravity.
Example 2:
Imagine that a rocket is launched a distance R away from the center of the planet of mass
Mp. What must its velocity be to escape the planets' gravitational pull?
This expressions gives you the minimum velocity required to overcome a planet's
gravitational pull. If the initial velocity is larger, the rocket will still escape and will have
some final kinetic energy when it is very far away from the planet.
PES 1110 Fall 2013, Spendier
Lecture 37/Page 6
Example 3:
What is the escape velocity of an object from the moon's surface?
Example 4:
If a rocket is launched from the surface of the moon with velocity vi = 3.00 km/s, what
velocity does it have very far away from the moon? Assume the rocket does not feel
gravitational pulls from other planets and the sun)