Download Worksheet 9: Gravitation

Worksheet 9: Gravitation
1. The only force that keeps planets (or other objects, like comets) in orbit around the
Sun is Newton's gravitational force.
a. Explain why the torque produced by this force about the Sun is zero.
The gravitational force is directed along the line that goes from the Sun to the planet,
so the lever arm r⟂ is zero and 𝜏 = F r⟂.
b. As a consequence of this, the angular momentum of a given planet with respect to
the Sun:
-
increases with r, the distance to the Sun.
decreases with r, the distance to the Sun.
is constant along the planet’s orbit (conserved).
Explain.
If the net torque is zero, angular momentum is conserved.
All planets and closed-orbit comets move along elliptical orbits, the Sun being at one
focus of the ellipse.
planet
Sun
The point where the distance of the planet to the Sun is maximum/minimum is called
aphelion/perihelion, respectively. Let vA/ vP and rA/ rP be the speed of the planet and its
distance to the Sun at the aphelion/perihelion.
(Worksheet 9)
c. Use your result from part b and the formula L = I ω, I = mr 2 , and ω = v / r to
prove that vA rA = vP rP
At the aphelion and perihelion, where the velocity of the planet is perpendicular
to the line between planet and the Sun, the magnitude of angular momentum is
L = Iω , with I = mr 2 and ω = v / r
where
I = moment of inertia of the planet around the Sun
m = mass of the planet
r = distance to the Sun
ω = angular speed of the planet at that point
v = tangential speed of the planet at that point
Using conservation of angular momentum,
(
I Aω A = I Pω P → mrA2
) vr = ( mr ) vr
A
A
2
P
P
→ rAv A = rP v P
P
d. From this we conclude, the speed of the planet relative to the Sun:
-
increases with r, the distance to the Sun.
decreases with r, the distance to the Sun.
is constant along the orbit (conserved).
Explain.
Angular momentum is proportional to the distance and the speed of the planet. If
the distance increases, speed must decrease, and vice versa.
(Worksheet 9)
2. Uranus has a radius of 25,560 km and a surface acceleration due to gravity of 8.86
m/s2. Its moon Miranda is in a circular orbit about Uranus at an altitude of 104,000 km
above the planet’s surface. Miranda has a mass of 6.6×1019 kg and a radius of 235 km.
Calculate:
a. The mass of Uranus
The acceleration of gravity on the surface of any planet is g =
the mass and radius of the planet. Thus
GM
where M and R are
R2
g U RU2 (8.86 m/s2 )(25.56 ×106 m) 2
MU =
=
= 8.68 ×1025 kg
−11
2
2
G
6.67 ×10 Nm /kg
b. The escape velocity on Uranus?
The escape velocity from the surface of Uranus is
vU = 2G
(
)(
MU
(8.68 × 1025 kg)
= 2 6.67 × 10−11 Nm 2 /kg 2
= 21.3 km/s
RU
25.6 × 106 m
)
c. The acceleration of Miranda for its motion about Uranus.
The distance between Miranda and the center of Uranus is
r = 104 ×106 m +25.56 ×106 m =129.56 ×106 m
Since the only force acting on Miranda is the gravitational pull from Uranus, the
acceleration of Miranda is the gravitational acceleration by Uranus at that distance
mM M U
= mM a
r2
M
8.68 ×1025 kg
a = g U (r ) = G 2U = 6.67 ×10−11 Nm 2 /kg 2
= 0.345 m/s 2
6
2
r
(129.56 ×10 m)
G
d. The acceleration due to Miranda’s gravity at the surface of Miranda.
19
GM M
kg
−11
2
2 6.6 ×10
gM =
= 6.67 ×10 Nm /kg
= 0.080 m/s 2
2
2
RM
(235000 m)
(Worksheet 9)
e. Describe the motion of an object released 1 m above Miranda’s surface, on the side of
Miranda that faces Uranus.
gU
gM
Miranda
Uranus
This question can be tricky. If we draw a figure, it is pretty clear that the net acceleration
of the object will be toward Uranus. But that does not mean that the object will really
“fall” toward Uranus (i.e., that it will get closer to Uranus).
The object will fall toward Miranda, just as one would initially expect. The object is
orbiting Uranus along with Miranda, and gU is exactly the right acceleration to account
for that part of the motion, just like it accounts for the orbiting motion of Miranda. The
part of the acceleration that comes from Miranda (gM) will pull the object back to
Miranda’s surface.
Note that this is the same situation that we have when an object is dropped near the
surface of the Earth, which is orbiting the Sun: the gravitational pull of the Sun also acts
on the object, and it is very important! It is what keeps the object moving with the Earth.
We usually do not worry about this acceleration because we look at the motion of the
object relative to the Earth (which we assume to be at rest, or at least an inertial
reference frame). This is an excellent approximation for any object whose motion is so
short that the motion of the Earth along its orbit can be taken to be rectilinear.