Answers to Problems
(show your work)
1. F = ma so a = F/m.
(a) 2F on m: a’ = F’/m’ = 2F/m = 2 (F/m) = 2a.
(b) 2F on 2m: a’ = F’/m’ = 2F/(2m) = (2/2) (F/m) = (1)(a) = a.
(c) 10F to m: a’ = F’/m’ = 10F/m = 10(F/m) = 10a.
(d) 10F to 3m: a’ = F’/m’ = 10F/3m = (10/3)(F/m) = 10/3 a = 3.33 a
2. This is just F = ma again, but we have to figure out a.
a = change in velocity / time = 2 m/s / 25 s = 0.08 m/s2.
F = ma = (2500 kg)(0.08 m/s2) = 200 Newtons
3. STUDENTS SHOULD DO THIS FOR THEMSELVES.
The weight of a body is just the gravitational force exerted on it. You can therefore find your
weight on the Earth or on the Moon from their surface gravities.
2
2
 The Earth’s surface gravity is 9.8 m/s while the Moon’s is 1.7 m/s .
 The Moon’s is thus 9.8/1.7 = about 5.8 times smaller than the Earth’s.
 Thus your weight on the Moon is your weight on the Earth divided by 5.8.
Alternatively, students can calculate the surface gravity of the Moon directly using g = GM/R2
and then their weight with F = mg; but they will have to come up with their mass in kilograms to
calculate the force in Newtons and/or convert that back to pounds or compare to their Earth
weight in Newtons (depending on instructor preference for the format of the answer).
4. From the problem, Neptune’s distance from the Sun is about 30 AU (more exactly it’s 30.05
AU). The orbital velocity, V, of a small mass around a much larger one can be found from the
1/2
formula in the chapter, namely, V = (GM/d) , where M is the Sun’s mass and d is Jupiter’s
distance from the Sun. Substituting:
-11
30
11
1/2
3
VJupiter = ( (6.7 × 10 m3 kg-1s-2 × 2 × 10 kg)/ (30 × 1.5 × 10 m) ) = 5.44 × 10 m/s.
The orbital period is the time it takes a body to complete an orbit.
Thus, orbital velocity = circumference/orbital period, V = C/P. Evaluating this we obtain
11
3
9
P = 2d/V = 2 × 30 × 1.5 × 10 m/ (5.44 × 10 m/s)= 5.2 × 10 seconds.
7
Since there are approximately 3.16 × 10 seconds in 1 year, P = about 165 years. Students should
be encouraged to check a result like this against the data in the appendix or Kepler’s 3rd law.
11
20
5. The Milky Way Galaxy has a mass of about 10 MSun, and the Sun is 2.6 × 10 m from the
center of the galaxy.
1/2
-11
30
11
20
V = (GM/R) = (6.7 × 10 m3 kg-1s-2 × 2 × 10 × 10 kg /2.6 × 10 m)1/2
5
V = 2.3 × 10 km/s
20
5
15
P = 2R/v = 2  × 2.6 × 10 /2.3 × 10 = 7.1 × 10 s.
One year is about 3.16 × 107 sec, so the period of the Sun’s orbit around the Milky Way in years
15
8
is 7.1 × 10 s /(3.16 × 107 s/yr) = 2.25 × 10 years = 225 million years.
6. The modified form of Kepler’s third law states that M = 4d3/GP2, where d is the orbital
radius (assuming it is circular) and P is the orbital period. In this problem d = 3 × 1011 meters
and P = 3 years. Before we can solve the problem, however, we need to convert P in years to P
in seconds.
7
124 d × 24 hr/1d × 60 min/1hr × 60 sec/1min = 1.07 × 10 sec.
Inserting the values for a and P, gives
M = 4 (5 × 1010 m)3/[6.67 × 10-11 m3 kg-1 s-2 (1.07 × 107 s )2]
= 4 × 125 × 1030-(-11)/[6.67 × (1.072 × 107x2 )] kg
= 4935/7.6 × 1041-14 kg
= 6.5 × 1029 kg
Since the Sun’s mass is 2 × 1030 kg, the mass of Gliese 581e is about 1/3 of a solar mass:
0.65 × 1030kg / (2 × 1030 kg/solar mass) = 0.33 solar masses.
7. In Section 3.7, the Earth’s surface gravity is compared with the Moon’s. We will need the
mass and radius of Jupiter and Pluto, which conveniently are given in terms of the Earth’s radius
and mass in the Appendix.
g Earth
g Jupiter
M Earth
GM Earth
1
2
M
REarth
11.19 2
125
Jupiter
317
.
9





 0.4
2
2
GM Jupiter 
317.9 317.9
1



R
 Earth 


2
RJupiter
11.19 
R


 Jupiter 
The surface gravity of Earth is about 4 tenths of the surface gravity on Jupiter; or Jupiter’s
surface gravity is about 2.5 times the Earth’s.
For Pluto,
gEarth
gJupiter
M Earth
1
2
M Pluto
0.0021  0.188  16.8


REarth 2  1 2 0.0021


 
RPluto  0.188 
The surface gravity of Earth is about 16.8 times as strong as the surface gravity of Pluto, which is
about 1/16.8 = 0.059 that of Earth.

1/2
8. The escape velocity for the Earth is found from: Vesc = (2GM/R)
Inserting the given values in the equation, we find that
1/2
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24
6
1/2
4
Vesc = (2GM/R) = (2 × 7 × 10 m3 kg-1 s-2 × 6 × 10 kg / 6 × 10 m) = 1.18 × 10 m/s
or Vesc = approx. 12 km/s
9. The conversion to mph helps put this speed into context for students.
11.8 km/s × (1 mile/1.609 km) × (3600 s/ 1h) = 26,400 miles/h = 26,400 mph
1/2
10. The escape velocity for the Sun may be found using the formula Vesc = (2GM/R)
with the
Sun’s mass and radius inserted. This gives
-11
30
8
1/2
Vesc = (2 × 6.7 × 10 m3 kg-1 s-2 × 1.989 × 10 kg/(6.95 × 10 m))
= 6.2 × 105 m/s = approx. 620 km/s
Or with the approximate values,
-11
30
8 1/2
5
Vesc = (2 × 7 × 10 × 2 × 10 /(7 × 10 )) = 6.3 × 10 m/s = approx. 630 km/s
11. To compare the escape velocity of Mars and Saturn, the method is similar to the previous
problem. The ratio simplifies to the expression,
VSaturn

VMars
 M Saturn  RMars 



 M Mars  RSaturn 
We then evaluate each expression using the appropriate mass and radius. From the appendix,
MMars = 0.1 MEarth, MSaturn = 95 MEarth, RMars = 0.5 REarth and RSaturn = 9.5 REarth.
Vesc(Saturn)/Vesc(Mars) = ( (95/0.1) x (0.5/9.5))
1/2
= ( (95/9.5) x (0.5/0.1) )1/2 = (10 x 5)1/2
= 7.1.
This is a good example of a question where the numbers can be easily estimated by writing out
the values and shifting terms around.
12. To calculate the ratio of the escape velocities from the Moon and Earth, start with the
formula for escape velocity:
1/2
1/2
Vesc-Moon = [2GMMoon/RMoon] and Vesc-Earth = [2GMEarth/REarth] .
Next, divide the Moon’s escape velocity by the Earth’s,
1
VMoon
VEarth
 2GM Moon  2


RMoon 



1
2
 2GM Earth 


 REarth 
M Moon
M
 R

M Earth
  Moon  Earth 
RMoon
 M Earth  RMoon 
REarth
We then evaluate each expression using the appropriate mass and radius; from the chapter, 81
MMoon = MEarth and REarth/RMoon = 3.8, so
VMoon
1
  3.8  0.22
VEarth
 81 
Vesc-Moon/Vesc-Earth = 0.22.
In other words, the escape velocity of the Moon is about 1/5 of the escape velocity of the Earth.
13. To find out if the pitcher can throw a ball fast enough to escape from Sinope, we need to
find the escape velocity of Sinope. To do that, we use the escape velocity formula, V =
(2GM/R)1/2.
Putting in the values for the mass and radius of Sinope, we find that
V = [2 × 6 × 1016 kg × 6.7 × 10-11 m3kg-1s-2/(1.8 × 104 m)]1/2
= [2 × 6 × 6.7 × 1016-11-4/1.8 (m/s)2]1/2
= [44.67 × 10]1/2 m/s
= 21 m/s
which is smaller than the speed of the pitch. Thus the ball can escape.