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course Phy201
Problem 8A tower begins at the surface of the Earth, at a distance of 6400 km from the center, and rises to a position 3000 kilometers further from the center. A person of mass 64 kg climbs the tower. The individual has an average power output of 1.01 watts / kg of body mass, and can sustain this output for 8 hours daily.
Approximate the work done against gravity for each of the first three 1000 kilometer segments of the tower.
How many days would be required to make the 3000 kilometer climb?
Solution
Earth's center to (6400 + 1000) km
Earth is 64 kg (9.8 m/s ^ 2) = 627.2 Newtons
radii will be (6400 + 1000) km / 6400 km
gravitational field (9.8 m/s ^ 2) / [(6400+ 1000)/6400] ^ 2
64 kg (9.8 m/s ^ 2) / [(6400+ 1000)/6400] ^ 2 = 469.14 Newtons.
The second segment extends from (6400 + 1000) km to (6400 + 2* 1000) km
64 kg (9.8 m/s ^ 2) / [(6400+2* 1000)/6400] ^ 2 = 272.335 Newtons
The third segment extends from (6400 + 2 * 1000) km to (6400 + 3 * 1000 km)
Problem 9
What is the acceleration of an object of mass 300 kilograms in circular orbit about a planet of mass 9 *10^ 24 kilograms at a radius of 20000 kilometers?
How fast must be object be traveling if it is to remain in a circular orbit about the planet?
How long will it take the object to complete one orbit? Neglect any difference between the radius of the orbit and the distance between the object and the center of the planet.
Solution
F = G m1 m2 / r ^ 2
gravitational force = 450.224 Newtons
gravitational acceleration = ( 450.224 Newtons)/ ( 300 kg) = 1.50074 meters/second ^ 2
v ^ 2 / r = v^2 / 2E+07 meters = 1.50074 meters/second ^ 2
v = 5478.5 meters/second
T = `ds / v = 1.25664E+08 meters/( 5478.5 meters/second) = 22937 seconds
Problem 10
We wish to determine the mass of a planet from the motion of a satellite. If our observations indicate that a small object orbits the planet at a distance of 20000 kilometers from the center of the planet with a period of 87 minutes, what is the mass of the planet?
Solution
87 minutes/5220 seconds
v = 24060 m/s.
Planet mass = M = ( 24060 m/s) ^ 2( 20000 m) / (6.67*10^-11 N m ^ 2/kg ^ 2)
= 173.5 * 10^24 kilograms
Problem 11
Imagine that you are orbiting a neutron star whose mass is 65 * 10^30 kilograms at a distance of 20 kilometers from its center.
Determine the average force gradient (force change per unit of distance), in Newtons per meter, for one kilogram masses between 20 and 20+1 kilometers from the center of the neutron star.
Use this gradient to estimate the difference you would therefore expect between one kilogram of your left shoulder and one kilogram of your right shoulder, assuming that one shoulder is .39 meter further from the star than the other.
Determine the average velocity gradient, in (m/s) per meter, between orbits at 20 km and at 20+1 km from the center.
Determine how long it would therefore take one shoulder to move ahead of the other by one complete revolution.
Solution
first distance the force is
1.084E+13 Newtons
second distance it is
9.832E+12 Newtons
difference for 1 km separation = 1.008E+12 Newtons
ave. force gradient = force / distance = 1.008E+12 Newtons / 1000 meters = 1.008E+09 Newtons/meter
Problem 12
What are the KE and PE changes and the ratio of PE to KE change if an Earth satellite of mass `mass kg, originally in a circular orbit of radius 7.3 * 10 ^ 6 m, increases its orbital radius to 7.49 * 10 ^ 6 m?
Solution
7.3 * 10 ^ 6 meters = ( 7.3 * 10 ^ 6 meters / (6.38 * 10 ^ 6 meters)^ -2 * 9.8 m/s^2 = `g1
7.49 * 10 ^ 6 meters = ( 7.49 * 10 ^ 6 meters / (6.38 * 10 ^ 6 meters) ^ -2 * 9.8 m/s^2 = 7.11
vel. at radius 7.3 = `sqrt ( 7.485 m/s^2 * 7.3 * 10 ^ 6 meters) = 7.391 * 10^3 m/s,
vel. at radius 7.49 = `sqrt ( 7.11 m/s^2 * 7.49 * 10 ^ 6 meters) = 7.297 * 10^3 m/s
KE change = ( 79860 - 81940) MJ = -2080 MJ
( 7.395 / rEarth) ^ -2 * 9.8 m/s^2 = 7.294 m/s^2
mass kg * 7.294 m/s^2 = 21880 N
distance parallel to field = ( 7.49 - 7.3) * 10 ^ 3 meters = .19 meters "