Open Query 28

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course Phy 121

028. `query 28

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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: 9.8=k/rE^2 and v^2=g(rE/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.

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).

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 Earty, 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): I don’t feel like I really understood this very well.

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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’=(6.67e-11*7.35e22)/1.74e6=1.619m/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 **

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Self-critique (if necessary): OK

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Self-critique rating #$&* OK

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

The derivative problem really confused me, but I am going to try to understand it better.

You don't need to understand the part about the derivative, just the result. You don't even have to remember that the result was associated with something called a derivative.

&#Your work looks good. See my notes. Let me know if you have any questions. &#

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