Does escape speed depend on launch angle? That is, if a
projectile is given an initial speed vo, is it more likely to
escape an airless, non-rotating planet, if fired straight up
than if fired at an angle?
A)  Yes
B)  No
C)  Impossible to tell from the information given.
The speed of escape depends only on the relationship between the
object’s KE and PE at launch, which are independent of angle. L31 W 11/5/14 1 1 2
mM
When thrust is turned off:
Etot = KE + PE = mv0 − G
2
R0
As long as the thrust is off, the rocket will have the same Etot. Its
distance R will change, but then so will its speed v.
(1) As R é, v ê. v0
R0 ≈ RE
RE
L31 W 11/5/14 2 1 2
mM
When thrust is turned off:
Etot = KE + PE = mv0 − G
2
R0
As long as the thrust is off, the rocket will have the same Etot. Its
distance R will change, but then so will its speed v.
(1)  As R é, v ê.
(2)  There is nothing in the conservation of energy about the angle at
which the rocket is fired. So, the angle doesn’t matter.
(3)  2GM
=
Etot = 0 Unbound Orbit Parabola v0 = vescape
R0
KE
is turned
when
thrustv0is<turned
Etot when
< 0 thrust
Bound
Orbit off
=
CPE
ircle,
Ellipse
vescape
off
Etot > 0 Unbound Orbit Hyperbola
v0 > vescape
L31 W 11/5/14 3 1 2
mM
When thrust is turned off:
Etot = KE + PE = mv0 − G
2
R0
As long as the thrust is off, the rocket will have the same Etot. Its
distance R will change, but then so will its speed v.
(1)  As R é, v ê.
(2)  There is nothing in the conservation of energy about the angle at
which the rocket is fired. So, the angle doesn’t matter.
(3)  2GM
=
Etot = 0 Unbound Orbit Parabola v0 = vescape
R0
Etot < 0 Bound Orbit Circle, Ellipse v0 < vescape
KE
is turned
off <
HPE
when thrust
Etot when
> 0 thrust
Unbound
Orbit
yperbola
v0is> turned
vescape
off
L31 W 11/5/14 4 1 2
mM
When thrust is turned off:
Etot = KE + PE = mv0 − G
2
R0
As long as the thrust is off, the rocket will have the same Etot. Its
distance R will change, but then so will its speed v.
(1)  As R é, v ê.
(2)  There is nothing in the conservation of energy about the angle at
which the rocket is fired. So, the angle doesn’t matter.
(3)  2GM
=
Etot = 0 Unbound Orbit Parabola v0 = vescape
R0
Etot < 0 Bound Orbit Circle, Ellipse v0 < vescape
Etot > 0 Unbound Orbit Hyperbola
v0 > vescape
KE when thrust is turned off > PE when thrust is turned off
L31 W 11/5/14 5 How about this: if a projectile is given an initial speed
vo, is it more likely to escape an airless but rotating
planet, if fired from the equator northward,
northeastward, or eastward?
It is most efficient to fire the
projectile toward the east.
Surface of the earth has an
eastward rotational speed, vrot.
vlaunch N
N
NE
E
vtot
E
vrot
L31 W 11/5/14 6 Suppose a projectile is fired straight upward from the surface of an
airless, nonrotating planet of radius R with the escape velocity vesc.
What is the projectile's speed when it is a distance 4R from the
planet's center (3R from the surface)?
Hint: remember for the escape velocity Etot = 0
A)  vesc B) vesc/2 C) vesc/3 D) vesc/4 E) vesc/9
1 2
mM
2GM
0 = Etot = mv − G
⇒v=
2
R
R
Therefore if R increases by a factor of 4, v reduces by a factor of 2.
L31 W 11/5/14 7 Assignments
Announcements:
•  HW 10 is due this week. CAPA 11 is now live.
•  You should read Ch. 10 now.
Today:
•  Finish up our discussion of orbits and gravity today. •  Will move on to rotational motion, found in Ch. 10. •  Also we’ll discuss center-of-mass on Friday, found in Ch. 9.
L31 W 11/5/14 8 To Create “Artificial Gravity” in Space
Simulate Earth: N = mg
centrifuge
r
period τ
x
What is the correct dynamical
relationship between the
centripetal force, N, and mg?
A)
B)
N=mg
L31 W 11/5/14 C)
D)
mv 2
=N
r
mv 2
= N − mg
r
mv 2
= N + mg
r
mv 2
=0
r
9 To Create “Artificial Gravity” in Space
Simulate Earth: N = mg
centrifuge
mv 2
= N = mg
r
r
period τ
x
⎛ circumference ⎞
v =⎜
⎟⎠
⎝
period
2
2
4π 2 r 2
=
= gr
2
τ
N=mg
L31 W 11/5/14 10 To Create “Artificial Gravity” in Space
Simulate Earth: N = mg
centrifuge
4π 2 r 2
= gr
2
τ
r
period τ
x
N=mg
To simulate earth’s gravity, at
what period must a wheel of
radius r rotate in space?
A)
r/g
B)
g/r
C) 2π r / g
L31 W 11/5/14 D) 2π g / r
11 To Create “Artificial Gravity” in Space
τ = 2π r / g
centrifuge
Let r = 10 m, then
r = 10 m
period τ
x
N=mg
L31 W 11/5/14 τ ≈ 2π sec ≈ 6.3sec
2π r 2π (10m)
v=
≈
τ
2π sec
≈ 10m / s
12 To Create “Artificial Gravity” in Space
The astronaut drops his pen and observes it
to 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.
Fnet = 0
L31 W 11/5/14 C) The pen falls because the artificial
gravity pulls it toward the floor.
13 To Create “Artificial Gravity” in Space
Example: https://www.youtube.com/watch?v=YQsp9upXH3Y
Spaceship Discovery 1
in the movie:
L31 W 11/5/14 14 Rotational Motion
-- rotation about a fixed axis.
As distinct from “translational motion” which we
have studied so far.
What you’ll see:
We’ll introduce variables to describe rotational motion (usually
using Greek letters) and their relationships to one another will
be exactly the same as for translational motion. The only
change will be the symbols used and the meaning, of course.
position velocity acceleration L31 W 11/5/14 Translational Motion
Rotational Motion
x (m) v (m/s) = dx/dt a (m/s2) = dv/dt θ (rad)
ω (rad/s) = dθ/dt
α (rad/s2) = dω/dt
15 Rotational Motion
-- rotation about a fixed axis.
As distinct from “translational motion” which we
have studied so far.
What you’ll see:
We’ll introduce variables to describe rotational motion (usually
using Greek letters) and their relationships to one another will
be exactly the same as for translational motion. The only
change will be the symbols used and the meaning, of course.
Constant acceleration a:
Constant angular acceleration α:
v = v0 + at
ω = ω 0 + αt
1 2
x = x0 + vot + at
2
v 2 = v02 + 2aΔx
1 2
θ = θ 0 + ω ot + α t
2
ω 2 = ω 02 + 2αΔθ
L31 W 11/5/14 16 Rotational Motion
-- rotation about a fixed axis.
As distinct from “translational motion” which we
have studied so far.
What you’ll see:
We’ll introduce variables to describe rotational motion (usually
using Greek letters) and their relationships to one another will
be exactly the same as for translational motion. The only
change will be the symbols used and the meaning, of course.
Translational Motion
Rotational Motion
And, as we’ll see, there are similar expressions for the
rotational version of mass, force, Newton’s laws, KE,
conservation of Energy, momentum and its conservation, and
so force.
L31 W 11/5/14 17 Rotational Kinematics
CCW: +
θ0 = 0
CW: -
Describing rotational motion
irrespective of rotational forces
s
Remember:
θ ≡
r
or
s = rθ
(θ: rads)
(r,θ) polar coordinates
Circumference = 2πr
360° = 2π rad
180° = π rad
L31 W 11/5/14 For circular motion, θ completely
describes the motion. It gives the
“rotational position” relative to a
reference θ0 = 0.
18