Query 28

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course Phy 201

7/12 around 3:00

8. `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:

 

 

a= k / r^2. When r = rE, a = 9.8 m/s^2 so

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

k = 9.8 m/s^2 * rE^2,

a= [ 9.8 m/s^2 * (rE)^2 ] / r^2.

a = 9.8 m/s^2 ( rE / 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 / r) and

v = sqrt( 2 g rE / r).

F = G m M / r^2,

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

v = sqrt( G M / r).

 

 

confidence rating #$&*:

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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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Self-critique (if necessary):

 

 

OK 

 

 

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Self-critique Rating: 3

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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:

 

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

 

 

confidence rating #$&*:

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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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Self-critique (if necessary):

 

 

 OK

 

 

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Self-critique Rating: 3

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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:

 

 

 I’m not sure?

 

 

confidence rating #$&*:

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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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Self-critique (if necessary):

 

 

 I see now. Once I worked it with the solution it made since :)

 

 

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