The Moon does not fall to Earth because A: It generates a

14-1
The Moon does not fall to Earth because
A: It generates a gravitational force equal and opposite the
earth's pull.
B: The net force on it is zero.
C: It is beyond the main pull of Earth’s gravity.
D: It is being pulled by the Sun and planets as well as by
Earth.
E: none of the above
Answer: E. None of these is correct. The moon DOES fall to earth, all the time!
There is a large force, F = GM(earth)M(moon)/R(earth to moon)^2 which pulls it
towards us. If this force was NOT there, the moon would travel in a straight line and
run away! A circular orbit means acceleration towards the center, that's what you
mean by "falling"! The reason the moon doesn't HIT the earth is because it has a
nice big SIDEWAYS velocity which never goes away (because it's always
perpendicular to the force of gravity). That means that as the moon falls towards us,
it also drifts sideways, and ends up further around the circle. No CLOSER to us
radially... You have to really think about this and make sense of it for yourself. The
moon IS accelerating towards you right now, but it keeps missing...
14-2
Suppose Earth had no atmosphere and a ball were fired
from the top of Mt. Everest in a direction tangent to the
ground.
If the initial speed were high (image)
enough to cause the ball to
travel in a circular trajectory
around Earth, the ball’s
acceleration would
A: be much less than g (because the ball doesn’t fall to
the ground).
B: be approximately g.
C: much greater than g
D: depend on the ball’s speed.
Answer: It's g. The acceleration of ANY object that is freely falling near the surface
of the earth is g. Sideways speed is irrelevant. This goes way back to the very
beginning of the class- it's Galileo's observation.
Air resistance would change it a wee bit. Also, "g" at th top of Mt. Everest is a
TINY TINY bit smaller than "g" here.
14-3
At some instant in time, two asteroids in deep space are a
distance r=20 km apart. Asteroid 2 has 10 times the mass
of asteroid 1. What is the ratio of their resulting
acceleration (due to gravitational attraction to each other),
m1
m2 = 10 m1
a1 / a2 =
A: 10:1
B: 1:1
D: Not enough information
C: 1:10
Answer: F(1 on 2) = F(2 on 1) (in magnitude), that's N-III.
(Or, use F = G M1 M2/r^2, it's the same if you reverse M1 and M2)
Since F=ma, and F is the same, the acceleration of the little one must be 10 times
bigger than the acceleration of the big one, 10:1.
14-4
Planet X has the same mass as earth, but its radius is only
half as big. How does gx compare with our g?
A:
B:
C:
D:
E:
gx = g
gx = 2g
gx = 4g
gx = (1/2) g
gx = (1/4) g
Answer: F(grav) = G M(planet) m(object) / R^2, = m(object) g,
so g = G M(planet)/R^2. Thus, if M is the same, and R is half, g(x) must be 4 times
bigger. (Same mass but smaller means planet X is very dense!)
14-5
The space shuttle orbits about 200 km above the earths
surface. The shuttle's "pre-launch weight" is its weight
measured on the ground. (Neglecting the fuel) The
magnitude of the force of gravity on the shuttle while it is
in orbit is
A:
B:
C:
D:
Slightly greater than pre-launch.
Equal to pre-launch
Slightly less than pre-launch
Lots less than pre-launch, almost (but
not quite) zero.
E: Precisely zero.
Answer: Slightly less than pre-launch. The force of gravity is nearly the same 200
km up, because G M(earth) M(ship)/R^2 depends on the distance R to the CENTER
of the earth (6000 km deep), not to the survace. Of course, 6200^2 is a bit bigger
than 6000^2, so there is a small difference. But it's not as much as you'd think.
The shuttle is not truly "weightless", it's just APPARENT weight that vanishes up
there. (It's in free-fall!)
14-6
An astronaut is floating around in the space shuttle's cabin.
Her acceleration, as measured from the earth's surface, is
A: zero - she's floating
B: very small, in some random direction
C: quite large, nearly g, directed towards
the center of the earth
D: quite large, nearly g, directed along the line of
travel of the shuttle
Answer: She's in orbit, so she's in uniform circular motion. She has a large
acceleration! We just said F(grav) is almost the same up there, so her acceleration is
almost the same as on earth in freefall, namely, nearly g, directed towards the center
of the earth. She's just like a ball that's been tossed up in the air, accelerating straight
down. The only difference is she has such a large sideways speed that when she has
fallen 200 km, she's gone so far to the side that the curved surface of the earth has
moved 200 km away from her, so she's still 200 km above the ground!!!
14-7
A rock is released from rest, in deep space, far from the
earth. At a certain point, as it falls towards us, it crosses the
moon's orbit.
moon
earth
rock
When the rock is the same distance from the earth as the
moon, the acceleration of the rock is
A: Greater than
B: Equal to
D: (Need more information)
the acceleration of the moon
C: Smaller than
Answer: The only force acting on either one is gravity,
F(grav) = G M(earth) m / R^2. The acceleration is then F/m, and that's the same for
both!!
As the rock falls toward the Earth, its acceleration is..
A: constant.
B: not constant.
Answer: as it approaches, R is getting smaller so GM/R^2 is getting bigger. The
acceleration is thus getting larger. It's not constant.
(It's only approximately constant when you're NEAR the surface, because R isn't
changing very much as you go from 6000km+0 to 6000 km + 20 feet, say. )
14--8
At time t=0, a satellite in circular orbit about the Earth is
directly over Denver, 300 miles above the city, traveling
east at 16,000 mph.
At the same time, a rock is released from rest 300 miles
above the city, very near the satellite.
rock
satellite
Denver
A: true or B: False
At t=0+, the accelerations of the rock and the satellite are
identical in magnitude and direction.
Answer: We've seen in this in several recent concept tests. The acceleration of both
are identical. (nearly g, towards the center of the earth). Both are freely falling, and
at the same radius.
Mass is irrelevant (it cancels from F=ma and F=G M m/r^2), and so is sideways
speed.
14-9
Two satellites A and B of the same
mass are going around Earth in
concentric orbits.
The distance of satellite B from Earth’s
center is twice that of satellite A.
What is the ratio of the magnitudes of
the accelerations of A to B?
B
A
|a(A)| / |a(B)| =
A: 8
D: 1
B: 4
C: 2
E: Sqrt(2)
Answer: a = G M(earth)/R^2, so if you're half the distance, you have 4 times the
acceleration.
14-10 A planet in elliptical orbit around a star moves from
the point in its orbit furthest from the star (A) to the closest
point (P). The work done by the force of gravity during
this movement is A: 0
B: +
C: A
P
Answer: You go faster when you're closer to the sun. Since W(net) = Delta K,
if you go faster, something has done + work on you. That something can only be
gravity, there are no other forces.
Another way to see it:
The force of gravity from the Sun is centripetal (toward the center). The force has a component
along the direction of displacement. W = F . Delta r = F * Delta r * cos(theta) > 0
Δr
FSun
The planet executes one complete orbit starting from point
A and returning to A. The work done by the force of
gravity during this orbit is:
A: 0
B: +
C: -
Answer: Zero. By the work energy theorem, or just conservation of energy.
14-11
Two satellites A and B of the same mass
are orbiting Earth.
The distance of satellite B from Earth’s
center is twice that of satellite A.
What is the ratio of the speed of A to
that of B?
|v(A)| / |v(B)| =
A: 1/2
D: Sqrt(2)
B: 1
C: 2
E: Sqrt(1/2)
Answer: F= ma, so GM(earth)m /R^2 = m v^2/R,
thus v = Sqrt[G M(earth)/R]. The farther you are, the slower you go.
So if A is 1/2 the radius, v is Sqrt[2] times bigger.
B
A
14-12
Two communications satellites are in orbit at the same
height, but one weighs twice as much as the other. The
speed of the heavier satellite is
A: Less than
B: Equal to
D: (need more information)
C: Greater than
the speed of the lighter one.
Answer:See previous problem, mass cancels out. Speed depends only on radius, for
spherical orbits.
14-13
Kepler's 3rd law: T^2/R^3=constant for all planets.
The period T for the earth is 1 year (!)
The distance of the earth to the sun is called 1 A.U.
Suppose an asteroid is in circular orbit around the sun at a
distance of 2 A.U. How long does it take the asteroid to
orbit the sun once?
A 2 years
B: 3 years
C: 2^(3/2) = 2.83 years
D: 2^(2/3) = 1.59 years
E: Not enough information (An asteroid is not a planet can't figure it out from Kepler's third law.)
Answer: T (period) goes like R^(3/2), so if you're 2 times as far, your year is 2^3/2
times longer.
14-14
Kepler's 3rd law: T^2/R^3=constant for all planets. The
period T for the earth is 1 year.
The distance of the earth to the sun is called 1 A.U.
Suppose a satellite orbits the earth at a distance of about
1/1000 AU. How long does it take the satellite to orbit the
earth once?
A:
B:
C:
D:
E:
Not enough information given.
1/1000 year.
1000 years.
(1/1000)^(3/2) years
(1/1000)^(2/3) years
Answer: Not enough information given. Kepler's law only works if the objects orbit
a common center. We need to know the earth's mass (or the distance to the moon, or
something) to figure this one out.
14-15
A rock, initially at rest with
3R
R
respect to Earth and located
far away is released and
accelerates toward Earth.
An observation tower is
built 3 Earth-radii high to observe the rock as it plummets
to Earth.
Neglecting friction, the rock’s speed when it hits the
ground is
A: twice
B: three times
C: four times
D: eight times
E: sixteen times
its speed at the top of the tower.
Answer: Conservation of energy says 1/2 mv^2 - G M(e) m/R = constant.
If it starts infinitely far away with v=0, then at "4R" it has
1/2 mv^2 = constant/(4R).
And at "R" it has
1/2 mv^2 = same constant/R, i.e. 4 times BIGGER KE.
So 2 times BIGGER v.
14-16
Does escape velocity depend on launch angle?
(That is, if a projectile is given an initial speed vo, is it
more likely to escape an (airless) planet, if fired straight up
than if fired at an angle?)
A: Yes
B: No
Answer: The formula for energy conservation only involves v^2, angle does not
matter.
14-17
A projectile is fired straight upward from the surface of an
airless planet (radius R) with escape velocity. What is the
projectile's speed when it is a distance 4R from the planet's
center (3R from the surface)?
A: 1/2 vesc
B: 1/4 vesc
C: 1/9 vesc
D: 1/3 vesc
E: None of these is correct.
(Ignore the gravity of the Sun and other astronomical
bodies.)
Answer: 1/2 m vesc^2 - G M m / R = 0 (that's the condition for escape velocity)
So later, 1/2 m v^2 - G M m / (4R) must still = 0. (energy is conserved)
So 1/2 m v^2 = (1/4) * 1/2 m vesc^2 is.
Therefore v^2 = (1/4) vesc^2.
or v = (1/2) vesc
14-18
An object moves along the x-axis . The potential energy
U(x) vs. position x is shown below.
U(x)
x
x=A
Wh
When the object is at x=A, which of the following
statements must be true?
A: The velocity is negative, i.e. along the -x direction.
B: The acceleration is negative.
C: The total energy is negative.
D: The kinetic energy is positive.
E: None of these statements is always true.
Answer: What we know is F = -dU(x)/dx, so the FORCE on the object at point A is
negative. (Therefore the acceleration is negative). We know nothing about the
velocity. The total energy could be anything, really. K could be zero, if this was a
"turnaround point". So it looks to me like only B is true.
14-19
A planet is in elliptical orbit around the Sun. The zero of
GMm
potential energy U is chosen at r = ∞, so U( r ) = − r .
How does the magnitude of U, i.e. |U| compare to the
KE? A: |U| > KE
B: |U| < KE
C: |U| =KE
D: depends on the position in the orbit.
Answer:The earth is bound, so K + U <0. (If it was =0 we'd have escape velocity!)
So K < -U = + G M m /r. Thus, K is LESS than |U|, or |U| > K. The numbers will
change at different points, but it must ALWAYS be true that K+U < 0 everywhere,
because that sum never changes. A diagram may help:
PE
r
Etot
PE KE
A satellite is in circular orbit around a planet that has a
very tenuous atmosphere extending up to the altitude of the
satellite. Due to atmospheric drag, the speed of the
satellite…
14-20
A: increases
B: decreases
C: remains constant
Answer: Very tricky. Drag will make the satellite's orbital radius drop, and since v =
Sqrt[ G M/r], the speed will INCREASE. How can this be? Well, K+U = total
energy, so K - GM m/R is the total energy. K is the smaller term (because you're in
orbit), the potential "wins". If R decreases, the negative term gets MORE negative,
the positive term gets MORE positive, but the total gets MORE negative (because
the negative one is winning). Your total energy is decreasing, your potential energy
is decreasing, your Kinetic energy is increasing (but not as fast). So they spiral in
faster and faster, burning up more and more energy and ultimately burning up
entirely!
14-21
Astronaut Dave Bowman is standing in the centrifuge of
the spaceship Discovery. He drops his pen and observes it
fall to the floor.
Which statement below is most accurate?
A: After he releases the pen, the net force on the pen is
zero.
B: The pen falls because the centrifugal force pulls it
toward the floor.
C: The pen falls because the artificial gravity pulls it
toward the floor.
Answer: A is correct. The pen was moving in a circule, and if he lets go, it will
henceforth follow a straight line path... which APPEARS to make it fall to the
ground in his reference from, but in fact there is no net force on it....