Query 28

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course PHY 231

11/21 11:12 am

028. `query 28 

PRELIMINARY STUDENT QUESTION:

Again I am still having trouble with what equations to use...for PE change we would use the equation PE = Gmm/r1 - 
Gmm/r2 same as the Force equation or do we just find dPE by grav. Force * ds both are listed in my notes. As far as 
KE I assume m*9.8m/s/s *rE/r^2

INSTRUCTOR RESPONSE

PE = -G M m / r^2 so going from radius r1 to radius r2 we would have `dPE = GMm/r1 - GMm/r2. 

If r1 and r2 don't differ by much, then G M m / r1 and G M m / r2 won't differ by much, and it would be important to use an appropriate number of significant figures in calculating the difference. 

Alternatively the difference can be expressed in terms of the common denominator r1 * r2 as G M m * ( r2 - r1) / (r2 * r1), where the common factor G M m has been factored out. 

As another alternative if r1 and r2 are proportionally close, you can figure out the average force and multiply by `dr = r2 - r1, as described below:

`dPE, i.e., the change in gravitational PE, is also equal to average gravitational force * change in distance from planet center:

• `dPE = F_ave * `dr

This is valid if `dr is small compared to r1 and r2, for the following reasons and with the associated cautions:

Gravitational force is not linearly related to r (i.e., the graph of F vs. r is not a straight line), so the average force on an interval is not equal to the average of the initial and final force. 

However if the proportional change in r is small enough the curvature of the graph won't be noticeable, and the average force could be approximated by the average of initial and final force.  In fact it is possible that the proportional change in r is small enough that over the relevant interval there is no significant change in the force itself.

• In these cases the change in PE is more easily found using F_ave * `dr.

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Question: `qQuery class notes #26

Explain how we use proportionality along with the radius rE of the Earth to determine the gravitational acceleration at distance r from the center of the Earth to obtain an expression for the gravitational acceleration at this distance. Explain how we use this expression and the fact that centripetal forces is equal to v^2 / r to obtain the velocity of a satellite in circular orbit.

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Your solution:

This is just like the question in the previous query where it's a proportional situation.

Force= m*(Re/r)^2*g.

Fcent= m*a_cent -----> a_cent = v^2/r -----> Fcent= m*v^2/r

Fcent=Fgrav -------> m*v^2/r=m*(Re/r)^2*9.8----> v=sqrt[(Re^2/r)*g]

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Given Solution:

`a** The proportionality is accel = k / r^2. When r = rE, accel = 9.8 m/s^2 so

 

9.8 m/s^2 = k / rE^2.

 

Thus k = 9.8 m/s^2 * rE^2, and the proportionality can now be written

 

accel = [ 9.8 m/s^2 * (rE)^2 ] / r^2. Rearranging this gives us

 

accel = 9.8 m/s^2 ( rE / r ) ^2, which we symbolize using g = 9.8 m/s^2 as

a = g rE^2 / r^2.

 

If we set the acceleration equal to v^2 / r, we obtain

 

v^2 / r = g ( rE / r)^2 so that

v^2 = g ( rE^2 / r) and

v = sqrt( g rE^2 / r) = rE sqrt( g / r)

Thus if we know the radius of the Earth and the acceleration of gravity at the surface we can calculate orbital velocities without knowing the universal gravitational constant G or the mass of the Earth.

 If we do know G and the mass of the Earth, we can proceed as follows:

The gravitational force on mass m at distance r from the center of the Earth is

F = G m M / r^2,

Where M is the mass of the Earth and m the mass of the satellite.  Setting this equal to the centripetal force m v^2 / r on the satellite we have

m v^2 / r = G m M / r^2, which we solve for v to get

• v = sqrt( G M / r).

**

 Self-critique (if necessary): Okay so I don't understand the expression for v above: v = sqrt( 2 g rE / r). What happened to Re^2 and where does the 2 come from? I looked over my expression and I don't see what's wrong with it, if the given one is actually the correct expression.

There's nothing wrong with your expression. The given solution was messed up.

 Self-critique Rating:2

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Question: `qUniv. 12.50 (12.44 10th edition). 25 kg, 100 kg initially 40 m apart, deep space. Both objects have identical radii of .20 m.

When 20 m apart what is the speed of each (relative to the initial common speed, we presume), and what is the velocity relative to one another? Where do they collide? Why does position of center of mass not change?

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Your solution:

m1 = 25 kg and m2 = 100 kg

Here I integrate Fgrav btw 40 and 20 m then use KE eq and Cons of M to find v's

v_m1= 1.63x10^-5 m/s

v_m2(20)=  4.08x10^-6 m/s

v1/2 = 1.22x10^-6 m/s

I don't understand how to find the distance where they collide.

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Given Solution:

`a

The question asks about the state of the system when the objects are 20 m apart.  This is because at the 20 m separations, the objects will first contact one another (each has diameter 20 m, so each has radius 10 m, and two sphere each with radius 10 m contact one another at the instant their centers are 20 m apart).

The question as phrased here didn't ask you to find the force or the acceleration at any point, but to put the problem in context we will do so below:

The forces at the r1 = 40 m and r2 = 20 m separations would be

• F1 = (6.67 * 10^-11 N m^2 / kg^2 * 25 kg * 100 kg) / (20 m)^2 = 1 * 10^-10 N

• F2 = (6.67 * 10^-11 N m^2 / kg^2 * 25 kg * 100 kg) / (20 m)^2 = 4 * 10^-10 N

Their accelerations are therefore

• accel of first mass at 40 m separation: a = F1 / m1 = 1 * 10^-10 N / (25 kg) = 4 * 10^-12 m/s^2

• accel of first mass at 20 m separation: a = F2 / m1 = 4 * 10^-10 N / (25 kg) = 16 * 10^-12 m/s^2 (written in this form for easy comparison with the first value, but more properly written 1.6 * 10^-11 m/s^2)

• accel of 2d mass at 40 m separation: a = F1 / m1 = 1 * 10^-10 N / (100 kg) = 1 * 10^-12 m/s^2

• accel of 2d  mass at 20 m separation: a = F2 / m1 = 4 * 10^-10 N / (100 kg) = 4 * 10^-12 m/s^2

** At separation r the force is F = G m1 m2 / r^2.

• For any small increment `dr of change in separation the approximate work done by the gravitational force is F `dr = G m1 m2 / r^2 * `dr.

• We take the sum of such contributions, between the given separations, to form an approximation to the total work done by the gravitational force.

• We then take the limit as `dr -> 0 and obtain the integral of G m1 m2 / r^2 with respect to r from separation r1 to separation r2.

• An antiderivative is - G m1 m2 / r.

• The change in the antiderivative between the separations r1 and r2 is

- G m1 m2 / r1 - (-G m1 m2 / r2) = G m1 m2 ( 1/r2 - 1 / r1).

• This expression is evaluated at r1 = 40 m and r2 = 20 m to get the change G m1 m2 ( 1/(20 m) - 1 / (40 m) ) in KE.

• We get KE of about 4 * 10^-9 Joules but you should verify that carefully.

We use conservation of momentum and kinetic energy to determine the final velocities:

Within a reference frame at rest with respect to initial state of the masses the initial momentum is zero.

• If the velocities at the 20 m separation are v1 and v2 the from conservation of momentum, we have

m1 v1 + m2 v2 = 0, so that v2 = -(m1 / m2) * v1.

The total KE, which we found above, is .5 m1 v1^2 + .5 m2 v2^2.

• Substituting v2 = - (m1 / m2) v1 and setting equal to the KE we can find v1, then from this we easily find v2.

• You might get something like 4.1 * 10^-6 m/s for the velocity of the 100 kg mass; this number is again not guaranteed so verify it yourself.

The position of the center of mass does not change because there is no external force acting on the 2-mass system.

• The center of mass is at position r with respect to m1 (taking m1 to be the 25 kg object) such that

m1 r - m2 (40 meters -r) = 0.

• Substituting m1 and m2 you get 25 r - 100 (40 meters - r ) = 0.

The solution for r is r = 4 / 5 * 40 meters = 32 m, approx.

This is the position of the collision relative to the initial position of the 25 kg mass.

This position is 8 meters from the 100 kg mass.

 Self-critique (if necessary): I don't know where that expression comes from: m1*r-m2(40-r) = 0

If you were to put m1 and m2 on opposite ends of a balance of length L, then if the fulcrum was placed at distance r from the mass m1, the moment arms would be r and L - r. Assuming m1 to be at the left end of the balance, the net torque would be m1 r - m2 ( L - r ). The condition for the fulcrum to be at the center of mass is that the net torque be zero.

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self-critique rating #$&*: "

Good. See my notes.

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