AST 105: Introduction to the Solar System
HOMEWORK # 2 Solutions
1. For objects that are in bound orbits, what is the major difference between Kepler’s First Law and Newton’s reformulation of it?
Kepler’s first law indicates that the Sun is at a focus. Newton’s reformulation is that the center of mass of the Sun-planet system is at the focus.
2. Approximately, Jupiter completes an orbit of the Sun every 12 years. Suppose
Jupiter had the same mass of the Sun. If its mean distance from the Sun were
to remain unchanged, what would be its new orbital period?
Newton’s orbital equation is
p2 =
4π 2
a3 .
G M⊙ + MJupiter
Jupiter’s mass is ordinarily much less than the Sun’s (M⊙ ), so we would have
p2 = Ka3
where K = 4π 2 /(GM⊙ ). If p is in Earth-years and a is in AU, then K = 1.
Now, if we increase Jupiter’s mass so that MJupiter = M⊙ , it should be obvious
from Newton’s equation that the new value for K is 1/2. If we don’t change
√
a, then we now have p2 = a3 /2 so that the new orbital period must be 12/ 2
years or about 8.4 years.
3. A rubber ball is aimed to strike at a perpindicular angle a flat board in space
and thrown against it. If it has a speed of 10 m/s when it hits the board, and
has the same speed when it rebounds, what is the change in velocity of the ball?
If the ball has a mass of 1 kg, and the board has a mass of 10 kg, what speed
does the board have after the impact?
This problem uses the Law of Conservation of Momentum. The total momentum before the ball impacts must be equal to the total momentum after
the ball impacts. The total momentum is mball vball + mboard vboard which is +10
m kg/s since the board is not moving and we count momentum in the ball’s
original direction as being positive (Remember, momentum is a vector and has
a magnitude and a direction). After the impact, the ball has the same speed
but it is directed in the opposite direction so mball vball = −10 m kg/s. Since
+10 m kg/s = mboard vboard − 10 m kg/s
we must have mboard vboard = +20 m kg/s. Since mboard = 10 kg, vboard = +2
m/s and the board is now travelling in the same direction as the ball did before
impact.
4. Which weighs more, a pound of lead or a pound of feathers?
Since we measure weight in pounds, they both weigh the same. Their
masses must also be the same, but their volumes are not.
5. Which has more volume, a pound of M & M’s on the Earth or a pound of M
& M’s on the Moon, or are their volumes the same?
A pound of M & M’s on the Earth must have less mass than a pound on
the Moon since weight = mass times gravity, and gravity on the Moon is much
less than on the Earth. Thus the number of M & M’s on the Moon is greater
than the number on the Earth, and the volume occupied by them on the Moon
must be greater than on the Earth.
6. Explain how the gravity (gravitational force) you experience depends on the
mass and radius of the planet you are on.
The gravity you experience is
g = GM/R2
since Force of gravity = mass times g.
What is the gravity of the Moon compared to the Earth? What is the gravity
of Mars compared to the Earth? Show your calculations. (Hint: look up the
masses and radii of the Earth, Mars and the Moon.)
Comparing gravity on the Earth compared to the Moon:
2
2
gEarth
GMEarth RM
MEarth RM oon
81.2
oon
=
=
= 6.03
=
2
gM oon
MM oon REarth
3.672
REarth GMM oon
Comparing gravity on the Earth compared to Mars:
2
2
GMEarth RM
MEarth RM ars
gEarth
9.35
ars
=
=
= 2.65
=
2
gM ars
MM ars REarth
1.882
REarth GMM ars
7. Explain how the escape velocity of a planet depends on its mass and radius.
The escape velocity is
s
2GMplanet
v=
.
Rplanet
What is the escape velocity of the Moon compared to the Earth? What is
the escape velocity of Mars compared to the Earth?
Comparing the escape velocity from the Earth compared to the Moon:
2GMEarth RM oon
MEarth RM oon
81.2
vEarth
=
=
= 22.1
=
vM oon
REarth 2GMM oon
MM oon REarth
3.67
Comparing the escape velocity from the Earth compared to Mars:
2GMEarth RM ars
MEarth RM ars
9.35
vEarth
=
=
= 4.97
=
vM ars
REarth 2GMM ars
MM ars REarth
1.88
8. The strength of tides depends, among other things, on the cube of the distance
between two objects. When the Moon formed, it is believed it was at a distance
from the Earth’s center of about 3 Earth radii. Today, the Moon is at a distance
from the Earth’s center of about 60 Earth radii. What was the strength of tides
from the Moon when the Moon was formed compared to their strength today?
If tides depends on the cube of the distance, and the Moon was 60 times
closer, the tides must have been 603 times as strong, or 216,000 times as strong.