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Homework 2
Author: Student, Amazing A.
September 26, 2016
Chapter 3
Problem 26
Earth is closer to the Sun in January than in July. Therefore, in accord with
Kepler’s second law: (a) Earth travels faster in its orbit around the Sun in July
than in January. (b) Earth travels faster in its orbit around the Sun in January
than in July. (c) It is summer in January and winter in July.
Kepler’s Second Law says that the planet sweeps through equal areas of the
orbit at equal times. As the Earth is closer to the Sun in January, it must move
faster then in order to satisfy this requirement. Therefore (b) is correct.
Problem 48
Eris Orbit: The recently discovered Eris, which is slightly larger than Pluto,
orbits the Sun every 557 years. What is its average distance (semi-major axis)
from the Sun? How does its average distance compare to that of Pluto?
Using p2 = a3 with the semi-major axis a in units of AU will give the period
p in years. If the orbit if 557 years long, then (557)2 = a3 = 310249. Therefore,
a = 67.7 AU. Pluto has a semi-major axis of 40 AU, nearly 60% smaller.
Extra Credit: Problem 50
Halley Orbit. Halley’s Comet orbits the Sun every 76.0 years and has an orbital
eccentricity of 0.97. a) Find its average distance from the Sun (semi-major
axis). b) Find its perihelion and aphelion distances.
To find the semi-major axis a, we can use the formula p2 = a3 (with units
of years and AU). If the period of the comet is 76.0 years, a = p2/3 , which gives
a = 17.9 AU.
The perihelion and aphelion distances are calculated by d = a(1 − e) and
d = a(1 + e), respectively. This gives dperi = 17.9(1 − 0.97) = 0.54 AU and
daphe = 17.9(1 + 0.97) = 35.3 AU.
1
Chapter 4
Problem 26
Suppose you visited another planet. (a) Your mass and weight would be the
same as they are on earth. (b) Your mass would be the same as on Earth, but
your weight would be different. (c) Your weight would be the same as on Earth,
but your mass would be different.
Weight is the gravitational force pulling your mass to the planet. Your mass
is invariant. Therefore, (b) is the correct choice. However, if the planet you
visited was identical to the Earth, your weight could be the same as well.
Problem 32
If Earth were twice as far from the Sun, the force of gravity attracting Earth
to the Sun would be (a) twice as strong. (b) half as strong. (c) one quarter as
strong.
The force of gravity can be calculated using We begin with the equation
Fg = G
M 1 M2
d2
(1)
If the masses remain constant, but the distance d is doubled
Fg ∝
1
1
= 2
d2
2
(2)
, and so doubling the distance between the Earth and the Sun will give a Fg of
(c) one quarter as strong.
Problem 51
Find the orbital period for the planet in each case. Hint: The calculations for
this problem are so simple that you will not need a calculator. (a) A planet with
twice the Earth’s mass orbiting at a distance of 1 AU from a star with the same
mass as the Sun. (b) A planet with the same mass as the Earth orbiting at a
distance of 1 AU from a star with four times the Sun’s mass.
We begin with the equation
P2 =
4π 2 a3
G(M1 + M2 )
(3)
Where a is the semi-major axis of the orbit and M1 , M2 refer to the two
orbiting bodies (Sun and Earth in this case). The problem states that we need
not use a calculator, and that a = 1 AU. We can simplify this by plugging in
the numbers for the Earth-Sun orbit, which has an orbital period of 1 year.
P2 =
4π 2 (1 AU )3
= 1 year
G(MSun + MEarth )
2
(4)
The mass of the Earth M ≈ 6 × 1024 kg is essentially negligible compared
to that of the Sun (M ≈ 2 × 1030 kg). We can rewrite the denominator into
Mtotal = MSun + MEarth ≈ MSun to make things easier. The rest of the terms
in the equation are constant. So, in units of years, we can write
s
1
P ∝
(5)
(Mtotal )
For problem (a), the book asks for a doubling of Earth’s mass. We know
that even by doubling the mass of the Earth, the planet’s mass will be less than
5 orders of magnitude smaller, and therefore Mt otal will remain constant. As
our period only depends on the combined mass of the system, which remains
unchanged, the period will not be significantly altered.
In (b), we are asked to compute the period of a system with a star that is
4 times more massive than our own Sun. So Mtotal = 4 × MSun + MEarth ≈
4 × MSun , and
s
1
= 0.5 years
(6)
P ∝
(4Mtotal )
By quadrupling the mass of the star the planet is orbiting, the period decreases
by half.
Chapter 5
Problem 28
Blue light has higher frequency than red light. Thus, blue light has (a) higher
energy and shorter wavelength. (b) Higher energy and longer wavelength than
red light. (c) Lower energy and shorter wavelength than red light.
The correct answer is (a). Blue light has more energy (E) than red light.
Frequency (ν) and wavelength (λ) are inversely proportional to one another, so
if blue light has higher frequency, it will have a lower wavelength and larger
energy.
hc
E=
= hν
(7)
λ
3