ASTR 1010 – Spring 2013 – Prof. Magnani
Answer Key – Homework 3
Question 1 – Ch. 3; #1
Empirical science is based primarily on observations and experimental data. It
primarily describes natural phenomena rather than explaining why these
phenomena occur. To do the latter, you need to devise theories based on
fundamental principles that often arise from logic, rational thought, and Occam’s
razor.
Question 2 – Ch. 3; #3
Nothing is at the other focus.
Question 3 – Ch. 3; #8
Kepler’s 3 Laws are empirical: they tell us how the planets move, but not why.
Newton’s Laws of motion and gravity are more general, apply to everything and
attempt to explain why Kepler’s Laws work in the context of more general,
fundamental principles of nature.
Question 4 – Ch. 3; #10
Inertia is the property that matter has that allows matter to resist acceleration.
Another way of saying it is that it is the property that matter has that allows it to
move at a constant velocity (recall that a constant velocity means there is no
acceleration).
Question 5 – Ch. 3; #12
For an object to deviate from straight-line motion, an acceleration must be provided.
In the case of circular motion, a centripetal acceleration must be provided to the
object. This acceleration is radial in direction and so does not change the object’s
tangential speed. Thus, there is an acceleration on the body.
Question 6 – Ch. 3; #16
Mass is the quantity of matter in an object and is measured in kilograms. Weight is a
force and so there must be an acceleration on the object for the weight to be defined.
As a force, the units of weight are Newtons (or kg m/s2).
Question 7 – Ch. 3; #17
We say Weight = mg (see formula on page 24).
But what is g?
g = GM/r2
clearly, M and r are the mass and radius of the body producing the gravitational
attraction that is felt as “weight”. The mass and radius for the Earth are different
from the Moon, so the constant of proportionality (g) is different on Earth than on
the Moon.
Question 8 – Ch. 3; #22
The comet on the elliptical orbit keeps coming back on a regular basis with a period
dictated by Kepler’s Third Law. The comet on a hyperbolic orbit swings by the Sun
once and then leaves the Solar System never to come back again.
Question 9 – Ch.3; #31
Remember that the force from gravity looks like this:
F = GM1M2/r2
So it decreases as 1/r2 . Since Uranus is 19 times farther from the Sun than Earth,
the gravitational force it will feel is 1/192 as strong. That is 1/361 as strong at
Uranus as at the Earth. So the Sun’s “grip” on the Earth is 361 times stronger than
on Uranus.
Question 10 – Ch. 3; #32
p 2 = a3
Here, avulc = 0.25 times Mercury’s distance from Sun = 0.25 * 0.387 = 0.0968 AU
So, p = 0.0301 years = 11 days, or 1/8 of Mercury’s period. You could have gotten
this right away, by realizing that if we write
pmerc2 = amerc3 then, Vulcan would be
pvulc2 = avulc3 = [(1/4) amerc]3 = [(1/64) amerc3 ]
pvulc = (1/8)pmerc
Question 11 – Ch. 3; #33
v = 2R/period = 2(4.5 x 109 km) / (164.8 years * 3.16 x 107 s/year) = 5.42 km/s
Question 12 – Ch.3; #35
MSun = V2R/G = 1.99 x 1030 kg
If you did not get this answer, did you convert the speed into m/s? Did you convert
the radius of the orbit into meters?
Question 13 – Ch.3; #36
Vesc = (2GM/R)0.5
M = 1 x 10-12 * Mearth = 5.97 x 1012 kg
R = 1 x 10-4 * Rearth = 6.378 x 102 m
Vesc = 1.12 m/s
So, with a good throw, you could throw it off the asteroid (the baseball would then
be orbiting the Sun – and following Kepler’s 3rd Law).
Question 14 – Ch. 3; #38
Acceleration at Earth’s surface = G Mearth/ Rearth2
Acceleration = 9.8 m/s2
Question 15 – Ch. 3; #42
Vesc = (2GM/R)0.5
Use the escape velocity formula to find the escape velocity from a point 500 km
above the Earth’s surface. Rearth = 6.378 x 106 m, so we need to add 500 km to that
to get Rpoint = 6.878 x 106 m
Plug that into the equation along with G and the mass of the Earth (5.97 x 1024 kg)
and you get an escape velocity of 1.076 x 104 m/s = 10.76 km/s
If the platform is already going 7.61 km/s, then all you need to do is launch your
spacecraft at 10.76 – 7.61 = 3.15 km/s in the direction that the platform is moving
and your spacecraft will be at the escape velocity for a point 500 km above the
surface of the Earth.
Question 16 – Ch.3; #45
M = 85 kg
Weight = Mg
gMars = 0.39(gearth) = 0.39 * 9.8 = 3.8 m/s2
Weight on Mars = 85 * 3.8 = 320 Nt.
Question 17 – Ch. 3; #47
Period of Zork, p = 7 yr. x 3.16 x 107 s/yr = 2.21 x 108 s
a = 7 AU x 1.5 x 1011 m/AU = 1.05 x 1012 m
Use Newton’s version of Kepler’s Third Law and solve for M:
M = 4 pi2 a3 /(G p2) = 1.4 x 1031 kg = 7 solar masses