Answer the following Review Questions from Nick Strobel’s AstronomyNotes: Chapter 5: Newton’s Law of Gravity
Review Questions
1.
What basic fundamental assumption did Newton make about the laws of nature on the Earth and in space?
Newton assumed that the same laws applied to Aristotle’s terrestrial matter and celestial matter; that is the laws of nature
applied equally to objects on the Earth and in the heavens. The law of gravity applied to a falling apple and to the orbiting
Moon. Newton’s triumph in formulating the Universal Law of Gravitation removed the distinctions between the terrestrial
matter and celestial matter of Aristotle and marked the official rejections of all things physical in Aristotle’s philosophy.
2.
Why is gravity often the most important force in astronomical interactions?
Gravity often the most important force in astronomical interactions because of (1) the exceedingly large masses of many
astronomical objects like planets, stars and galaxies that make the gravitational fields around those objects strong and long
ranged and (2) the overall neutrality of matter electrically so that the electric force is unimportant.
3.
What things does gravity depend on?
Gravity is an attractive force between any two bodies. The strength of gravity depends on two things.
1. The strength of gravity is directly proportional to the mass of both objects. If you increase the mass of either of the
objects in the problem, the gravitational force will increase. i.e. FGravity  M 1 M 2
2.
The strength of gravity follows an inverse square relation with distance. If you increase the distance between the two
bodies the gravitational force between the two bodies will decrease as the square of the distance increase. i.e.
FGravity 
4.
1
d2
How does gravity vary with distance between objects and with respect to what do you measure the distances?
The strength of gravity varies as one over the distance between the two bodies squared. The distance is measured from the
center of each body.
5.
What would happen to the Earth’s orbit if the Sun suddenly turned into a black hole (of the same mass)? Why?
If the Sun were to collapse into a black hole the same mass as the Sun, the Gravity field around the Sun would not change
and the Earth and other planets would continue in their regular orbits. This is because the strength of gravity around the Sun
depends only the mass of the Sun not its radius or density.
6.
What important laws of planet motion can be derived from Newton’s law of gravity?
All three of Kepler’s Laws of planetary motion can be derived from Newton’s Law of Gravity.
7.
Why is gravity called an “inverse square force”? What is the difference between a simple inverse relation and an inverse
square relation?
As expressed in Wikipedia, “In physics, an inverse-square law is any physical law stating that a specified physical quantity
or intensity is inversely proportional to the square of the distance from the source of that physical quantity. In equation form:
The divergence of a vector field which is the resultant of radial inverse-square law fields
with respect to one or more sources is everywhere proportional to the strength of the local
sources, and hence zero outside sources. Newton's law of universal gravitation follows an
inverse-square law, as do the effects of electric, magnetic, light, sound, and radiation
phenomena.”
As illustrated in the graph below, the Inverse Square Relation looks similar to the Inverse
Relation except that the Inverse Square Relation weakens more rapidly as distance increases.
4
3.5
Inverse Relation
3
Inverse Square Relation
2.5
2
1.5
1
0.5
0
0
8.
0.5
1
1.5
2
2.5
3
3.5
4
If the Earth was 3 A.U. from the Sun (instead of 1 A.U.), would the gravity force between the Earth and the Sun be less or
more than it is now? By how many times?
The gravity force between the Earth and the Sun be less than it is now by 9 times?
9.
If Mercury was 0.2 A.U. from the Sun (instead of 0.4 A.U.), would the gravity force between Mercury and the Sun be less or
more than it is now? By how many times?
The gravity force between the Mercury and the Sun be more than it is now by 4 times?
10. Why do astronauts in orbit around the Earth feel “weightless” even though the Earth’s gravity is still very much present?
The astronauts in orbit around the Earth feel “weightless” even though the Earth’s gravity is still very much present because
they are in free fall. Much as water in a cup with a hole in the bottom stays in the cup as the cup is falling, thus giving the
water the appearance of weightlessness, so the astronauts are falling together with their spacecraft and appear to have not
weight. The difference between the falling cup of water and the astronauts is that the astronauts have a very large tangential
velocity that carries them around the curvature of the Earth as they fall. For a very good visual of this effect see the
Newton’s Mountain website here.
11. What keeps satellites orbiting the Earth moving along their curved paths?
Satellites keep orbiting the Earth moving along their curved paths because of their very large tangential or horizontal
velocities. As the satellites fall due to gravity, their large tangential velocity carries them around the curvature of the Earth,
thus then never get any closer to the Earth even though they are in freefall.
12. What two things must be determined first in order to calculate the mass of a planet or a star?
4 2
 R 3 an equation for the mass of the central object can be
From Kepler’s 3 Law in its general form P 
GM Jupiter
2
rd
derived that looks like M Jupiter

4 2 R 3

. So you see that you need to know the orbital radius of a satellite (planet
G P2
around a star or a moon around a planet) and the orbital period of the satellite to determine the mass of the central object.
13. Jupiter’s moon Io has about the same mass as the Moon and orbits Jupiter at about the same distance that the Moon orbits the
Earth (center to center). Then why does Io take only 1.8 days to orbit Jupiter but our Moon takes 27.3 days to orbit the Earth?
According to Kepler’s 3rd Law in general form According to Kepler’s 3 rd Law in general form P 
2
4 2
 R 3 relates
GM Jupiter
the periods of Jupiter’s moons to the radii of their orbits.relates the period of a moon to the radius of its orbit. Since R the
radii of the Moon and Io’s orbit are about equal, but P the periods of the two satellites are very different, then the difference
in P must be due to the difference in the MCentral the mass of each central body (i.e. Earth and Jupiter). In fact using ratios one
can write that
M Jupiter
M Earth
2
2
P
  27.3 d 
P
 Moon2   Moon   
  230
P
1
.8
d
PIo


 Io 
2
So that the mass of Jupiter is a couple to a few hundred times the mass of the Earth.
14. Astronomers were able to accurately measure the orbital periods of the moons of Jupiter since the time of Galileo, so why
was an accurate value for Jupiter’s mass not found for over 300 years until the astronomical unit was measured accurately?
According to Kepler’s 3rd Law in general form P 
2
4 2
 R 3 relates the periods of Jupiter’s moons to the radii of
GM Jupiter
their orbits. If both P and R are known for a moon, then the mass of Jupiter can be calculated by rearranging the formula to
solve for MJupiter. The problem that early astronomers had was that they did not know the actual distance in meters or
kilometers of the Earth to the Sun. Without precise quantitative knowledge of the size of the AU, the radii of Jupiter’s moons
could not be ascertained. Ground-based observations of Jupiter’s moons reveal only the angular separation of the moons
from Jupiter. To determine the actual distance in meters of the moons from Jupiter you need to know the distance between
Earth and Jupiter at the time of observation. Thus, even though the periods of the Jupiter’s moons were well known, their
orbital radii were not known until the size of the AU was determined during the early part of the 18 th century.
15. Which would have a shorter orbital period, a planet orbiting a massive star at 3 A.U. or a planet orbiting a low-mass star at 3
A.U.? Explain your answer.
Both the orbits at three AU will have the same circumference, so the planet that is moving faster will have the shorter orbital
period. The orbital speed of the planet is given by
vOrbit 
GM Star
where MStar is the mass of the star and R is 3 AU. You
R
can see from the equation that the higher the mass of the star the faster the orbital speed for a constant orbital radius. Thus
the planet orbiting the massive star will have a shorter orbital period because it will have a higher orbital velocity compared
to a planet orbiting a low mass star at the same distance.
16. If a planet orbiting a massive star has the same orbital period as a planet orbiting a low-mass star, which of the planets orbits
at a greater distance from its star? Explain your answer.
This problem is very similar to the previous problem. We need to know that the orbital speed of a planet around a star is
given by
vOrbit 
GM Star
where MStar is the mass of the star and R is the orbital radius of the planet. If a planet were
R
orbiting a massive and a low mass star at the same distance , the planet around the low mass star would take longer to
complete its orbit because its orbital velocity would be lower due to the lower mass of the star. So, we would need to move
the planet around the low mass star inward, so that R decreased and the orbital velocity increased to shorten its orbital period.
Thus, the planet around the massive star will be farther from away compared to the planet around the low mass star if tey are
to have the same orbital period.
17. How can you predict the orbital period of Jupiter’s satellite Europa
from observations of the other jovian moon Io?
As we demonstrated in class, you can use Kepler’s 3rd Law in its
general form in a very clever and simple way to learn about one
moon of a planet from knowledge of another moon. Let’s review
that demonstration. Imagine Io and Europa orbiting Jupiter as
shown in the figure to the right. Write Kepler’s 3 rd law for each
moon and then create the ratio:
2
PEuropa

4 2
4 2
3
2
 REuropa
 RIo3
and PIo 
GM Jup
GM Jup
Europa
REuropa
Io
RIo
Jupiter
2
PEuropa
PIo2

4 2
3
 REuropa
GM Jup
4 2
 RIo3
GM Jup
2
or

3
REuropa
RIo3
 PEuropa   REuropa 

  

P
R
 Io   Io 
3
So from the last equation above you can see that if the ratio of orbital radii of two moons are known then the ratio of orbital
periods can be determined or vice versa. We’ll apply this concept in the next problem.
18. If Io takes 1.8 days to orbit Jupiter at a distance of 422,000 kilometers from its center, find out how long it would take Europa
to orbit Jupiter at 671,000 kilometers from its center.
2
3
 PEuropa   REuropa 
  
 . If we fill in values for the parameters
Continuing our work from the previous problem we have 
P
R
 Io   Io 
in the equation we get
2
 PEuropa   671,000 km 

  

 1.8 d   422,000 km 
3
2
 PEuropa 

  1.59 3
 1.8 d 
2
 PEuropa 

  4.02
 1.8 d 
PEuropa
 2.005
1.8 d
PEuropa  2.005  1.8 d  3.6 days
So we find that Europa will take 3.6 days to orbit Jupiter.
19. If the Moon were twice as far from the Earth as it is now, how long could a solar eclipse last? (Solar eclipses currently last up
to about two hours from the start of the cover-up to when the Moon no longer blocks the Sun at all.)
The key to solving this problem (approximately) is to recognize that if the Moon were moved farther away from the Earth its
orbital velocity would decrease. It would slow down. Further, if the Moon’s orbital velocity decreases then it will take
longer to appear to pass inform of the Sun. So the duration of the eclipse will increase. Let’s quantity these.
Let us assume that the Moon currently orbits the Earth at a distance of R0 and that the orbital velocity of the Moon is given by
vOrbit 
GM Earth
. We don’t need to figure out the numbers. Now assume that we move the Moon so that it orbits the
R0
Earth at twice the distance 2R0. What will the Moon’s new orbital velocity be compared to the original orbital velocity?
Well, write the express for the new orbital velocity
vNew Orbit 
GM Earth
1 GM Earth
1



 vOrbit  0.71  vOrbit .
2  R0
2
R0
2
So the new orbital velocity of the Moon will be about 71% of the original orbital velocity. Thus the new orbital velocity is
29% slower than the original orbital velocity. Since the orbital velocity is 29% slower, the eclipse should take 29% longer to
complete. Thus a normal solar eclipse that lasts 2 hours will take 2 hrs
 1.29 = 2.58 hrs = 2 hrs 35 min.
GSS
RGSS
20. The Hubble Space Telescope orbits the Earth 220 kilometers above the
surface and takes about 1.5 hours to complete one orbit. How can you
find out how far up to put a communication satellite, so that it takes 24
hours to circle the Earth? (Such an orbit is called a “geosychronous
orbit” because the satellite remains above a fixed point on the Earth.)
RHST = 220 km + 6,378 km = 6,598 km
220 km
HST
=
6378 km
This problem is very similar to Problem 18 and we will use the variant
P
of Kepler’s 3rd Law  GSS
 PHST
2
  RGSS
  
  RHST
3

 where GSS stands for

Earth
geosynchronous satellite and HST is the the Hubble Space Telescope.
The only trick in this problem is to recognize that we are not given the
radius of the Hubble Space Telescope’s orbit – only its height above the
Earth’s Surface. See the figure to the right. You can see, I hope, that the
radius of the HST orbit is given by 220 km plus the radius of the Earth
or 220 km + 6,378 km = 6,598 km. No we solve the problem just like
Problem 18.
 PGSS

 PHST
2
  RGSS 
  

  RHST 
3
2
 24 hr   RGSS 


 
 1.5 hr   6,598 km 
3
3
 RGSS 

  16 2
 6,598 km 
3
 RGSS 

  256
 6,598 km 
1
RGSS
 256 3
6,598 km
1
RGSS  6,598 km  256 3
RGSS  41,895 km
So the geosynchronous satellite must orbit the Earth with an orbital radius of 41,895 km.
21. Find out how long it will take the Cassini spacecraft to travel to Saturn 9.5 A.U. from the Sun.
Consider the figure to the below. A spacecraft leaves Earth orbit 1 AU from the Sun and travels to Saturn 9.5 AU from the
Sun on an orbit known as the Hohmann Transfer Orbit. The Hohmann Transfer Orbit has a major axis of 1 AU + 9.5 AU =
10.5 AU. Thus the semi-major axis a is 5.25 AU. Kepler’s 3rd Law states the orbital period of a spacecraft on an orbit with a
= 5.25 AU is given by P = a3/2. Thus the orbital period on this transfer orbit is (5.25) 3/2  12 years. Since the spacecraft is
only going to travel on half the orbit (from Earth to Saturn and not on to return) the cruise time to Saturn will be half the
orbital period or about 6 years.
9.5 AU
Hohmann Transfer Orbit
 
1 AU