PHY 151 Homework Solutions 13
Name: _____________________
Gravitational Potential Energy
1.
A satellite in Earth orbit has a mass of 100 kg and is at an altitude of 2.00 x 106 m.
What is the potential energy of the satellite-Earth system?
r is the distance from the center of the Earth to the satellites orbit.
U = -GMEm/r = -(6.67 x 10-11)(5.97 x 1024)(100)/(6.37 x 106 + 2.00 x 106)
U = -4.76 x 109 J
2.
How much work is done by the Moon’s gravitational field on a 1000 kg meteor as
it comes in from outer space and impacts on the Moon’s surface?
W = U = -(-GMmm/r – 0) =
r is radius of the moon
W = -(-6.67 x 10-11)(7.36 x 1022)(1000)/1.74 x 106 = 2.82 x 109 J
3.
At the Earth’s surface, a projectile is launched straight up at a speed of 10,000
m/s. To what height will it rise? Ignore air resistance and the rotation of the Earth.
Kf + Uf = Ki + Ui
0 – GMEm/rf = ½ mv2 – GMEm/ri
– GME/rf = ½v2 – GME/ri
-(6.67 x 10-11 x 5.98 x 1024)/rf = ½ (10000)2 – (6.67 x 10-11)(5.98 x 1024)/6.37 x 106
-4.0 x 1014/rf = 5 x 107 – 6.26 x 107 = -1.26 x 107
rf = 4 x 1014 / 1.26 x 107 = 3.17 x 107
height = (3.1.7 – 6.37) x 106 = 2.53 x 107 m
Energy Consideration in Planetary and Satellite Motion
4.
A space probe is fired as projectile from the Earth’s surface with an initial speed
of 2.00 x 104 m/s. What will its speed be when it is very far from Earth? Ignore
atmospheric friction and the rotation of the Earth.
Kf + Uf = Ki + Ui
½ V2 = ½ v2 – GME/r
V2 = v2 – 2GME/r
V2 = (2 x 104)2 – (2)(6.67 x 10-11)(5.98 x 1024)/6.37 x 106
V2 = (2 x 104)2 – (2)(6.67 x 10-11)(5.98 x 1024)/6.37 x 106
V2 = (4x 108) - (2)(6.67 x 10-11)(5.98 x 1024)/6.37 x 106
V2 = (4x 108) – 1.25 x 108 = 2.75 x 108
V = 1.66 x 104 m/s
5:
A 1000 kg satellite orbits the Earth at a constant altitude of 100,000 m.
(a)
How much energy must be added to the system to move the satellite into a
circular orbit with altitude 200,000 m?
(b)
What is the change in the system’s kinetic energy
(c)
What is the change in the system’s potential energy
(a)
Low orbit
Energy of satellite is –GMm/2r = -6.67x10-11 x 5.97x1024 x 1000/2(6.37x106 + 100000)
E = -3.0773 x 1010 J
High orbit
Energy of satellite is –GMm/2r = -6.67x10-11 x 5.97x1024 x 1000/2(6.37x106 + 200000)
E = -3.0304 x 1010 J
E = (-3.0304 – (3.0773) x 1010 = 4.69 x 108 J
(b)
K = -E = -4.69 x 108 J
(c)
U = 2E = 9.38 x 108 J