Query 28

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course PHY 201

028. `query 28PRELIMINARY 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:

Gravitational Field Strength equals Earth's Radius over orbital radius, squared and then mutliplied by gravitational acceleration(g).

g * (Re/r)^2

F = mg

F = (1/4)mg (at 2 Re)

F = (1/8)mg (at 3 Re)

F = (1/16)mg (at 4 Re)

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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).

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Question: `qPrinciples of Physics and Gen Phy problem 5.30 accel of gravity on Moon where radius is 1.74 * 10^6 m and mass 7.35 * 10^22 kg.

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Your solution:

To find the acceleration of gravity on the moon we will use the formula:

g = G(m/r^2)

g = (6.67 * 10^-11)(7.35 * 10^22 kg) / (1.74 * 10^6)^2

g = 1.619 m/s^2

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Given Solution:

`a** The acceleration due to gravity on the Moon is found using the equation

g' = G (Mass of Moon)/ radius of moon ^2

g' = (6.67 x 10^-11 N*m^2/kg^2)(7.35 X 10^22 kg) / (1.74 X 10^6 m)^2 = 1.619 m/s^2 **

STUDENT COMMENT

The problem is set up correct but was not solved for. The answer for the acceleration from gravity is 1.619 m/s^2

INSTRUCTOR RESPONSE

the numbers work out to 6.67 * 7.35 / (1.74)^2 * 10^-11 * 10^22 / (10^12), with 10^12 being the square of 10^6.

The powers of 10 therefore work out to 10^(-11 + 22 - 12) = 10^(-1).

You shouldn't require a calculator to get this must.

You can then use the calculator to find 6.67 * 7.35 / (1.74)^2.

your result is 16 * 10^-1 = 1.6. The units work out as indicated.

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Question: `qQuery gen phy problem 5.40 force due to planets (Mv, Mj, Ms, are .815, 318, 95.1 Me; orb radii 108, 150, 778, 1430 million km).

What is the total force on Earth due to the planets, assuming perfect alignment?

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Your solution:

We must find the magnitude of gravitational force between the objects using the formula:

F = G(m1 * m2) / r^2

For Mv:

F = (6.67 * 10^-11)(5.98 * 10^24)(4.8737 * 10^24) / (4.2 *10^10)^2

F = 1.1 * 10^18

Mj:

F = (6.67 * 10^-11)(5.98 * 10^24)(1.9 * 10^27) / (6.28*10^11)^2

F = 1.92 * 10^18

Ms:

F = (6.67 * 10^-11)(5.98 * 10^24)(5.67 * 10^26) / (6.52 * 10^11)^2

F = 5.32 * 10^17

Total Force = Me - Mv + Mj + Ms

TF = 0 - 1.1 * 10^18 + 1.92 * 10^18 + 1.38 * 10^17

TF = 9.58 * 10^17

Confidence rating: 1 (had help with problem)

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Given Solution:

`a** Using F = G m1 m2 / r^2 we get

Force due to Venus: F = 6.67 * 10^-11 N m^2 / kg^2 * (5.97 * 10^24 kg) * (.815 * 5.97 * 10^24 kg) / (1.5 * 10^11 m - 1.08 * 10^11 m)^2 = 1.1 * 10^18 N, approx.

Force due to Jupiter: F = 6.67 * 10^-11 N m^2 / kg^2 * (5.97 * 10^24 kg) * (318 * 5.97 * 10^24 kg) / (1.5 * 10^11 m - 7.78 * 10^11 m)^2 = 1.9 * 10^18 N, approx.

Force due to Saturn: F = 6.67 * 10^-11 N m^2 / kg^2 * (5.97 * 10^24 kg) * (95.7 * 5.97 * 10^24 kg) / (1.5 * 10^11 m - 1.43 * 10^11 m)^2 = 1.4 * 10^17 N, approx.

Venus being 'inside' the Earth's orbit pulls in the direction of the Sun while Jupiter and Saturn pull in the opposite direction so the net force is

-1.1 * 10^18 N + 1.9 * 10^18 N + 1.4 * 10^17 N = .9 * 10^18 N = 9 * 10^17 N, approx.. **

STUDENT QUESTION

Where do we get 1.5 * 10^11 m as part of r? I’m slightly confused as to where that came from?

INSTRUCTOR RESPONSE

Each calculation will be based on the masses of Earth and the planet in question, and on the distance between them when they are perfectly aligned.

The distance between Earth and another planet at perfect alignment is the difference of their distances from the Sun.

For example Venus is 1.08 * 10^11 m from the Sun, while Earth is 1.5 * 10^11 m from the Sun. The distance between Earth and Venus, when they line on a straight line with the Sun (and on the same side of the Sun) is therefore 1.5 * 10^11 m - 1.08 * 10^11 m.

Thus the quantity (1.5 * 10^11 m - 1.08 * 10^11 m)^2 in the denominator of the calculation for Venus.

Calculations for Jupiter and Saturn are similar, but each uses the appropriate distance of that planet from the Sun, and of course the mass of the planet.

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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:

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Given Solution:

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.

STUDENT QUESTION

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

INSTRUCTOR RESPONSE

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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Question: `qQuery gen phy problem 5.50 24 m diam wheel, rot period 12.5 s, fractional change in apparent weight at top and at bottom.

What is the fractional change in apparent weight at the top and that the bottom of the Ferris wheel?

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Your solution:

v = 2 * pi * r / T

v = 2 * pi * 12 / 12.5

v = 6.03m/s

a = v^2 / r

a = (6.03m/s)^2 / 12

a = 3.03m/s^2

Since the force and acceleration point inward,

Top:

F + Fg = w

ma + mg = w

m(a + g) = w

m(3.03m/s^2 + 9.8m/s^2) = w

m(12.83m/s^2) = w

Bottom:

F - Fg = w

ma - mg = w

m(a - g) = w

m(3.03m/s^2 - 9.8m/s^2) = w

m(6.77m/s^2) = w

ratio of weight = top/bottom

12.83 / 6.77 = 1.895/1

Confidence rating: 2

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Given Solution:

`a** Centripetal acceleration is a = v^2 / r.

For a point on the rim of the wheel, v = dist in 1 rev / time for 1 rev = `pi * 24 m / (12.5 sec) = 1.9 pi m/s, approx. (about 6 m/s)

Thus v^2 / r = (`pi * 1.9 m/s)^2 / 12 m = 3 m/s^2, approx.

At the top the only accel is the centripetal, and it is acting toward the center, therefore downward. The forces acting on any mass at the top are the gravitational force and the force exerted by the wheel on the mass. At the top of the wheel the latter force is the apparent weight. Thus

grav force + apparent weight = centripetal force

- m * 9.8 m/s^2 + wtApparent = m * (-3 m/s^2 )

wtApparent = m (-3 m/s^2) + m ( 9.8 m/s^2) = m (6.8 m/s^2).

A similar analysis at the bottom, where the centripetal force will be toward the center, therefore upward, gives us

- m * 9.8 m/s^2 + wtApparent = m * (+3 m/s^2 )

wtApparent = m (+3 m/s^2) + m ( 9.8 m/s^2) = m (12.8 m/s^2).

The ratio of weights is thus 12.8 / 6.8, approx. **

A more elegant solution obtains the centripetal force for this situation symbolically:

Centripetal accel is v^2 / r. Since for a point on the rim we have

v = `pi * diam / period = `pi * 2 * r / period, we obtain

aCent = v^2 / r = [ 4 `pi^2 r^2 / period^2 ] / r = 4 `pi^2 r / period^2.

For the present case r = 12 meters and period is 12.5 sec so

aCent = 4 `pi^2 * 12 m / (12.5 sec)^2 = 3 m/s^2, approx.

This gives the same results as before. **

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Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment.

STUDENT QUESTION

So if you have 2 spheres in space and they both start at rest, they will start moving toward each other because of the

gravitational attraction. Is the reason that doesn’t happen on Earth because of friction and the fact that the attraction is

so tiny compared to the pull of the earth?

INSTRUCTOR RESPONSE

Suppose you have two 1 kg spheres, with their centers separated by 10 cm. The gravitational attraction between the spheres would be about 10^-10 as great as the gravitational attraction of each to the Earth.

So for example if one of the masses was suspended by a thread 1 meter long, then if the other mass is brought to within .1 meter, the position of the suspended mass would change by about 10^-10 meters, the approximate diameter of a single atom. We wouldn't be able to measure that.

the gravitational forces between objects on the Earth are just too insignificant to be measured in this way.

Now we could put the system into a vacuum and allow the suspended mass to oscillate back and forth past the other mass. The gravitational attraction would affect the period of the pendulum, and over a large number of oscillations we might be able to detect the effect.

If we modify this idea slightly and use a torsion pendulum, then we have the basic idea of the Cavendish balance (you can search that term, which I believe you will encounter fairly soon in your text).

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Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment.

STUDENT QUESTION

So if you have 2 spheres in space and they both start at rest, they will start moving toward each other because of the

gravitational attraction. Is the reason that doesn’t happen on Earth because of friction and the fact that the attraction is

so tiny compared to the pull of the earth?

INSTRUCTOR RESPONSE

the gravitational forces between objects on the Earth are just too insignificant to be measured in this way.

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Query 28

&#This looks good. Let me know if you have any questions. &#