Download Homework 1. Estimate the speed of the planet mercury? Compare

Homework
1. Estimate the speed of the planet mercury? Compare your speed to mach one, and the
speed of light. (The period is 88 days, and the (average) orbital radius 0.387 AU )
Solution:
• The orbital period is 88 days and the radius is R = 0.4 AU = 0.4 × 1.5 × 108 km
so the speed is
(2πR)
= 49.5 km/s
v=
88days
So the speed of mercury is quite a bit faster than the speed of sound
v
' 140
(330 m/s)
and quite a bit slower than light
v
1
'
c
6400.
2. In a given time ∆t which is short compared to the orbital period of the moon (27 days),
the moon will fall a distance
1
am ∆t2
2
towards the center of the earth, where am is the acceleration of the moon. am is
analogous to the free fall acceleration g = 10 m/s2 on earth. What is the ratio between
g = 10 m/s2 and am ? How did Newton use this number to understand the distance
dependence of the gravitational force?
Solution:
• Using the result we derived in class for circular motion
am =
2
vm
Rem
we substitute the velocity of the moon, and the earth moon distance Rem =
384, 000 km = 3.84 × 108 m
vm = (2πRem )/(27days) = 1.034 km/s = 1034 m/s
and then we find
am =
Or in terms of g we have
(1034 m/s)2
= 0.0028 m/s2
3.84 × 108 m
am
' 0.00028
10 m/s2
1
• Newton noticed (cleverly) that this ratio of accelerations is just about equal to
the ratio of inverse radii.
Re 2
) = 0.00024
(
Rem
which to about 15% is equal to the number given above. Newton interperted this
as saying that the force of gravity decreases as
Fg ∝
1
R2
3. At all moments in time the earth is falling towards the sun with an acceleration aES .
How large is this acceleration in units of g (the acceleration of falling bodies on earth).
Why don’t we fall into the sun? Assume that the earth suddenly stopped moving.
Then, assuming (unrealistically) that our acceleration towards the sun is constant
thereafter, how long would it take us to reach the sun? Solution:
• The acceleration is
ve2
Res
We do not fall into the sun because we are moving so each time we fall we also
move forward leading to a circular motion.
Calculating the velocity of the earth in we have, for the earth sun radius Res =
1.5 × 1011 m and 1y ' π × 107 s
a=
ve =
(2πRes )
= 30 km/s
1 year
(1)
which is somewhat slower than the mercury speed of 47 km/s. Substituting numbers into the v 2 /R formula
a=
(30, 000 m/s)2
= 6 × 10−3 m/s2
1.5 × 1011 m
Or in units of g = 10 m/s2 we have
a
= 6 × 10−4
g
• Well then, the acceleration due to gravity would be the same as computed above
(i.e. the gravitational force is the same, so the acceleration is the same. Then if
we fell towards the sun with constant acceleration we would find
1
∆x = at2
2
So since ∆x = 1 AU = 1.5 × 1011 m we can determine t.
1.5 × 1011 m = 0.5 (6 × 10−3 m/s2 ) t2
2
=⇒
t ' 7 × 106 s ' 81 days
4. If a bullet is shot horizontally (by a person), estimate how far it travels before reaching
the ground. To answer this question you will need to estimate the muzzle speed of a
bullet and the height of a person1 .
Solution:
• Ok. First lets estimate that the height of a person (from foot to shoulder) is
h = 1.5 m. The horizontal motion is independent of the vertical motion (Galileo).
So the time it takes to fall is just
1
h = gt2
2
Or then
(1.5 m) = 0.5(10 m/s2 )t2
=⇒ t = 0.54 s
During this time, the bullet moves forward with a constant velocity in the forward
(or x) direction, v0x = 1/4 mach1. Of course the bullet is getting faster all the time
in the z direction, but this does not affect the motion in the forward direction.
Mach one is 330 m/s so this speed is v0x = 82 m/s. Thus the distance the bullet
travels is
∆x = v0x ∆t = 82 m/s (0.54 s) = 45 m
5. A bullet is shot straight up – estimate its initial speed.
(a) Sketch its velocity as a function of time. Label the time axis (i.e. the x-axis) in
appropriate units. Label the y axis (velocity) in appropriate units.
(b) When does it reach the top of the arc? Indicate this point on your graph.
(c) Use your graph to determine the average speed of the bullet on its way up.
(d) How high up did the bullet travel?
Solution:
• OK, starting from before we take vup = 80 m/s
(a) OK, the graph is shown below . We start with 80 m/s up. Every second
our speed decreases by 10 m/s so after 8 seconds our upward speed is 0 m/s.
Meaning we are no longer moving up.
1
Oh well . . . Ok . . . , I’ll help you out on the bullet question. . . A bullet is moved by exploding gasses, which move at the speed of sound in
air (mach one). So perhaps a bullet moves at about 1/4 of mach one.
3
80
Upward velocity (m/s)
60
average speed during upward flight
40
20
top of arc
0
-20
-40
-60
-80
0
2
4
6
8
10 12 14 16
time (s)
The negative velocities mean that we are moving down.
(b) On its way up was between time 0 and 8 s. The average speed during this
time is just halfway between 80 m/s (the start) and 0 m/s. This is shown on
the graph.
(c) Well, since the average speed during the upward flight was 40 m/s during the
first 8 s the distance travelled is
∆x = v̄∆t = (40 m/s)(8 s) = 320 m
4
(2)