Download Cunningham, Drew – Homework 30 – Due: Apr 14 2006

Cunningham, Drew – Homework 30 – Due: Apr 14 2006, 4:00 am – Inst: Florin
This print-out should have 11 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering. The due time is Central
time.
001 (part 1 of 2) 10 points
Given: G = 6.67259 × 10−11 N m2 /kg2 .
A 517 kg geosynchronous satellite orbits a
planet similar to Earth at a radius 198000 km
from the planet’s center. Its angular speed at
this radius is the same as the rotational speed
of the Earth, and so they appear stationary
in the sky. That is, the period of the satellite
is 24 h .
What is the force acting on this satellite?
Correct answer: 541.362 N.
Explanation:
4 π2
(6.67259 × 10−11 N m2 /kg2 ) (86400 s)2
× (1.98 × 108 m)3
= 6.15224 × 1026 kg .
=
003 (part 1 of 3) 10 points
C
The density of a sphere is given by ρ(r) = .
r
The sphere has a radius of 2.53 m and a mass
of 1011 kg.
Find the constant C.
Correct answer: 25.138 kg/m2 .
Explanation:
The mass of a differential element of the
sphere is
Solution:
2πr
T
v2
r
4 π2 r
= Msatellite
T2
4 π 2 (1.98 × 108 m)
= (517 kg)
(86400 s)2
= 541.362 N .
4 π2 3
r
GT2
r = 2.53 m and
M = 1011 kg .
Let : G = 6.67259 × 10−11 N m2 /kg2 ,
Msatellite = 517 kg , and
T = 86400 s .
v=
Mplanet =
1
dm = ρ dV =
C
(4 π r2 dr) .
r
Z
Z
F = Msatellite
(1)
002 (part 2 of 2) 10 points
What is the mass of this planet?
Correct answer: 6.15224 × 1026 kg.
Explanation:
Using the general gravitation law and Eq.
1, we have
Msatellite Mplanet
F =G
r2
4 π2 r
= Msatellite
T2
Msatellite Mplanet
4 π2 r
= Msatellite
, so
G
r2
T2
M=
r
dm = 4 π C
R dR ,
0
¶ ¯r
µ
1 2 ¯¯
= 4π C
R ¯
2
0
= 2 π C r2
= (12.8018 m2 ) π C
M
C=
(12.8018 m2 ) π
1011 kg
= 25.138 kg/m2 .
=
(12.8018 m2 ) π
004 (part 2 of 3) 10 points
Find the gravitational field 3.84 m from the
center of the sphere.
Correct answer: 4.57492 × 10−9 N/kg.
Explanation:
Let :
r = 3.84 m .
Cunningham, Drew – Homework 30 – Due: Apr 14 2006, 4:00 am – Inst: Florin
g=
(6.6726 × 10−11 N · m2 /kg2 ) (1011 kg)
(3.84 m)2
= 4.57492 × 10−9 N/kg .
005 (part 3 of 3) 10 points
Find the gravitational field 1.43 m from the
centre of the sphere.
Correct answer: 1.05391 × 10−8 N/kg.
Explanation:
Let : R = 1.43 m .
µ ¶
R
8. ac = g
r
µ ¶2
R
1
9. ac = g
2
r
Explanation:
Basic Concepts: The Universal Law of
Gravity is
m1 m2
F =G
.
r2
Gravitational potential energy is
The gravitational field inside the sphere is
GM
R2
2 π G C R2
=
R2
= 2πC G
= 2π(25.138 kg/m2 )
× (6.6726 × 10−11 N · m2 /kg2 )
g=
= 1.05391 × 10−8 N/kg .
006 (part 1 of 2) 10 points
A satellite with mass m is orbiting around
the Earth on a circular path with a radius
r. Denote the mass and the radius of the
Earth by M and R, respectively, and the
gravitational acceleration at the surface of
the Earth by g.
The magnitude of the centripetal acceleration of the satellite is given by
1. ac = g
g
2. ac =
2
³ r ´2
R
3. None of these
µ ¶
1
R
4. ac = g
2
r
µ ¶2
R
5. ac = g
correct
r
6. ac = g
1 ³ r ´2
7. ac = g
2
R
2
U = −G
m1 m2
r
and if we set U = 0 at r = ∞.
Solution: The force of gravity acts as centripetal force, Fc = FG , or
m ac = G
Mm
r2
so
GM
.
(1)
r2
One might already recall the expression for g;
however, if we don’t, for an object of mass m0
on the surface of the Earth
ac =
m0 g = G
so that
g=
M m0
R2
GM
.
R2
(2)
(The actual value, 9.8 m/s2 , differs somewhat
from this value due to other effects, such as
the rotation of the Earth).
Therefore our ac is
GM
G M R2
ac = 2 =
=g
r
R2 r 2
µ ¶2
R
.
r
(3)
007 (part 2 of 2) 10 points
The minimum increment of energy needed for
the satellite to escape (i.e., to leave its orbit
and move to infinity) is
1. ∆E =
GM m
r
Cunningham, Drew – Homework 30 – Due: Apr 14 2006, 4:00 am – Inst: Florin
GM m
correct
2. ∆E =
2r
GM m
3. ∆E =
4r
GM m
4. ∆E =
3r
GM mR
5. ∆E =
r2
2GM m
6. ∆E =
3r
GM mr
7. ∆E =
4 R2
G m2
8. ∆E =
4r
G M2
9. ∆E =
3r
3GM m
10. ∆E =
4r
Explanation:
If we insert the expression
ac =
v2
r
into our force equilibrium equation (1) in Part
1, we can solve for the velocity squared v 2
v2 =
GM
r
and consequently the kinetic energy of the
satellite is
K=
1
GM m
m v2 =
.
2
2r
Also, the potential energy is
U =−
GM m
,
r
so the total energy of the satellite is
E =K +U
GM m GM m
=
−
2r
r
GM m
=−
.
2r
Note: This could also be seen from readymade formulae.
3
Now we just have to remember that we set
the potential energy at infinite distance to be
zero. The least energy with which an object
escapes the gravitational attraction is when it
has no velocity after escaping. Thus, we need
to bring E up to zero. To do this we must add
an increment of energy ∆E so that
E + ∆E = 0 .
Therefore
∆E = −E =
GM m
.
2r
008 (part 1 of 3) 10 points
Given: G = 6.67259 × 10−11 N m2 /kg2
After a supernova explosion, a star may undergo a gravitational collapse to an extremely
dense state known as a neutron star, in which
all the electrons and protons are squeezed together to form neutrons. A neutron star having a mass of 1.7 × 1030 kg (about equal to
that of the Sun) would have a radius of about
9.62 km.
Find the free-fall acceleration at its surface.
Correct answer: 1.22573 × 1012 m/s2 .
Explanation:
The free-fall acceleration at the surface of
the neutron star is given by
GM
r2
= (6.67259 × 10−11 N m2 /kg2 )
(1.7 × 1030 kg)
×
(9620 m)2
= 1.22573 × 1012 m/s2 .
g=
009 (part 2 of 3) 10 points
Find the weight of a 83.9 kg person at its
surface.
Correct answer: 1.02838 × 1014 N.
Explanation:
The weight of the person is given by
w = mg
= (83.9 kg) (1.22573 × 1012 m/s2 )
= 1.02838 × 1014 N .
Cunningham, Drew – Homework 30 – Due: Apr 14 2006, 4:00 am – Inst: Florin
4
Ei = Ef (neglecting air friction), so
010 (part 3 of 3) 10 points
Find the energy required to remove a particle
of mass 1.65 × 10−27 kg from its surface to
infinity.
Correct answer: 1.94559 × 10−11 J.
Explanation:
The energy required to remove the particle
from the surface of the star to infinity is the
negative of the potential energy of the particle
at the surface, so we have
G M Mp
E=
r
= (6.67259 × 10−11 N m2 /kg2 )
(1.7 × 1030 kg) (1.65 × 10−27 kg)
×
(9620 m)
−11
= 1.94559 × 10
J.
011 (part 1 of 1) 10 points
Given: G = 6.67259 × 10−11 N m2 /kg2 .
A projectile is fired vertically upward from
the surface of planet Moorun of mass 3 ×
1024 kg and radius 3 × 106 m.
If this projectile is to rise to a maximum
height above the surface of Moorun equal to
6 × 106 m, what must be the initial speed of
the projectile?
Correct answer: 9.43228 km/s.
Explanation:
Basic Concept: Total energy (Kinetic
and Gravitational Potential) is
E=
GM m
1
m v2 −
.
2
r
Solution: Denote the mass of the projectile
by m. The initial total energy (before takeoff) is
1
GM m
Ei = m v 2 −
.
2
R
The final total energy (when the projectile is
at the maximum height) is
Ef = 0 −
GM m
,
R+h
since the projectile has zero velocity at its
highest point. Energy conservation implies
1
GM m
GM m
m v2 −
=−
.
2
R
R+h
Solving for v, we have
s
2M h
v= G
R (R + h)
½
= (6.67259 × 10−11 N m2 /kg2 )
2 (3 × 1024 kg) (6 × 106 m)h
×
(3 × 106 m) (9 × 106 m)
= 9432.28 m/s
= 9.43228 km/s ,
(1)
¾ 12
where
R + h = (3 × 106 m) + (6 × 106 m)
= 9 × 106 m .
Comment: Notice that if we had applied
ordinary kinematics, we would have utilized
the familiar formula
v12 − v02 = 2 a (x1 − x0 ) ,
which in our problem, is
0 − v 2 = 2 (−g)(h − 0) ,
so that the initial velocity to rise to height h
would be
p
v = 2gh
(2)
q
= 2 (9.8 m2 /s2 ) (6 × 106 m)
= 10.8444 km/s ,
which is NOT the right answer. This is
because the attraction of gravity decreases as
we go further away from the planet, requiring
less initial velocity than if the attraction had
been constant (which our simple kinematics
formula assumes). It is, however, a good
approximation when we are close to a planet’s
surface.
Try picking a smaller height h and see when
the simpler description (2) presents a good
approximation to the velocity of formula (1) !