Assignment 28 Query

Good. See my note(s).

Let me know if there's anything you need me to explain further.

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19:39:32 Query 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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RESPONSE --> centripetal acceleration v2 / r is equal to the gravitational field strength g = 9.8 m/s2 * (rE / r) 2 at distance r from the center of the earth

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19:39:43 ** 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 ( r / rE ) ^2. **

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RESPONSE --> ok

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19:39:46 23:58:06

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RESPONSE -->

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19:39:48 23:58:27

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RESPONSE -->

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19:39:52 23:59:46

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RESPONSE --> ok

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19:39:57 00:11:56

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RESPONSE --> ok

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19:52:21 Gen Phy 5.36 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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RESPONSE --> F = G * m1 * M2 / r^2; and assuming that the mass of the earth is Solving the F for each of these planets since the force is aligned it would be all in one direction I dont think I completely understand the variable that are listed... I have 3 sets of values of forces for four planets.. My first reaction is as what I started above and solve for the forces that each planet creates...but then I realize that I do not have the masses for each...It seems that I am missing a variable... that I need either the mass of the acceleration of gravity on the planets to solve. I was going to use the above equation and then add up the Fnet the the planets create....but it seems that the forces are already liste..but the values I would think to be too low to be the actual force..but this is my stabe after this Fnet = .815 + 318 + 95.1 = 413 Newtons ?? I am sure this is wrong

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19:59:25 ** 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.. **

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RESPONSE --> I must have been mising some data in the previous problem?!? --- Also this does not match up to problem 5.36 that I did in the book.. sorry for my confusion.... Looking back at my homework this was actually problem number 5.40 ... of which I understand this a little more. The values were the mass .. I cant believe I didnt remember this problem since it took so long to solve. But in any case the logic that I have previously would correctly solve with the exception that I did not equate out the negative foces for those plantes in the suns orbit.... I made the total forces of the plantes a total force to the earth

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20:11:27 Gen Phy 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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RESPONSE --> v = 2 pi r / T = 2 * 3.14 * 12/ 15.4 = 4.89 m/s The velocity would be approx. 4.9 which is half of gravity...

4.9 m/s is not an acceleration, so is not a quantity that could be compared with the acceleration of gravity.

v^2 / r would be the centripetal acceleration.

Also the denominator according to the information given above would be 12.5 s, not 15.4 sec, so v would be closer to 6 m/s.

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20:11:34 00:19:19

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RESPONSE --> ok

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20:16:07 ** 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 m/s, approx. 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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RESPONSE --> I will go back and work this problem out again...I seems that I failed to apply some key concepts...

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20:16:12 00:20:55

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RESPONSE --> ok

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20:16:16 Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE --> ok

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