1. A satellite is placed in a circular Earth orbit with radius R. A second satellite is placed in a circular
orbit of radius 2R. Compared with the first satellite, the second will have
a)
b)
c)
d)
e)
Less energy and less angular momentum
More energy and less angular momentum
Less energy and more angular momentum
More energy and more angular momentum
The same energy and the same angular momentum
2. Consider the elliptical orbit (shown below). At points A and B where the object is closest and
furthest from M, its velocity vector is perpendicular to the line connecting each point to the
center of M, labeled RA and RB. If the speed of the object at A is vA,, what will be the speed at B?
m
A
RA
M
RB
B
VA
3.
In March 1999, the Mars Global Surveyor (GS) entered its final orbit about Mars, sending data
back to Earth. The orbit the GS entered was slightly elliptical with its closest approach to Mars at
3.71 x 105 m above the surface and its furthest distance at 4.36 x 105 m above the surface. If the
speed of the GS at closest approach is 3.40 x 103 m/s, calculate the speed at the furthest point of
the orbit. (The mass of the GS is 930 kg and the radius of Mars is 3.43 x 106 m)
1.
Answer: D
More energy and more angular momentum
For circular orbits E = - G m M/2R so increasing R makes E less negative, or larger.
The angular momentum of the first satellite is L = mv R = m (GM/R) R = m (GMR)
So the second satellite with R2 = 2R1, will have a bigger angular momentum by a factor of 2
2.
LA = LB
m vA RA = m vB RB
so vB = RA vA / RB
So the object will slow down as it moves from A to B
3.
Ms v1 r1 = ms v2 r2 so v2 = v1 r1 / r2
= v1 (Rc + RM) / (RF + RM)
Where Rc and RM are the distances of closest and furthest approaches, respectively, and RM
is the radius of Mars.
V2 = ( 3.40 x 103 m/s ) (3.71 x 105 m + 34.3 x 105 m)/ (4.36 x 105 m + 34.3 x 105 m)
V2 = 3.34 x 103 m/s