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CHAPTER 11
THE GRAVITATIONAL FIELD
• Newton’s Law of Gravitation • Inertial mass and gravitational mass • Gravitational potential energy
• Satellites
• Kepler’s 3rd Law
• Escape speed
Newton’s Law of Gravitation
A force of attraction
occurs between two
masses given by
A value for G, the universal gravitational constant, can be
obtained from measurements by Henry Cavendish (1731 –
1810) of the specific gravity of the Earth. He found that
the density of the Earth is 5.48 times the density of water,
Cavendish’s measurements allowed the first determination
of the mass of the Earth; it is within 1% of today’s
accepted value.
r m1 m2
i.e., ρEarth = 5.48×1000 kg ⋅m−3 = 5480 kg ⋅m−3.
∴mE = volume × density = 43πRE
3 ρE
= 43π(6.40 ×106m)3 × (5480 kg ⋅m−3) = 6.02 ×1024 kg.
F = G
m1m2
r2.
mE mA
r = h + RE
The force on the apple (i.e.,
its weight) is
When released the acceleration of the apple is
Very close to the Earth’s surface, and
where is the Earth’s radius. We know
and
From above, the acceleration due to gravity a distance r
from a mass m is:
a =
FEAmA
= GmE
(h + RE )2.
RE >> h a ! g.
∴G =
RE2
mEg,
RE g = 9.81 m/s2,
mE = 6.02 ×1024 kg RE = 6.4 ×106 m,
∴G = 6.67 ×10−11 N ⋅m2 /kg2.
FEA = mAg(r) = GmEmA
r2
= GmEmA
(h + RE )2.
g(r) = G m
r2.
This expression provides us a way of determining the
acceleration of gravity at the surface of a massive object
like a planet. If the mass is and the radius is then
Examples:
• Mars:
• Moon:
In deriving these expressions we have assumed that the
Earth and planets are spherical. In fact, the Earth is really
a “flattened” sphere … look …
gp = Gmp
R2p
.
mp
Rp
mass = 6.42 ×1023 kgradius = 3.39 ×103 m
gMars = 3.7 m/s2
mass = 7.35×1022 kgradius = 1.74 ×103 m
gMoon = 1.6 m/s2
Req
RP
For the Earth
In fact, Since ones
weight is mg, one weighs less
• up a mountain,
• in an airplane,
than at sea-level!
At 30,000 ft:
Req > Rp.
g = G
mE
RE2
∴geq < gP.
geq ~ 9.83 m/s2 and gP ~ 9.78 m/s2.
Δww
~ −0.3%.
Gravitational and inertial mass
So far, without stating it explicitly we have used the same “mass” (m) in Newton’s Law of gravitation as was defined by Newton’s 2nd Law in chapter 4. Note that in the definition of the inertial mass of an object,
inertial mass = minert =
Fa
,
there is no mention of gravity. The mass used in Newton’s Law of gravitation is called the gravitational mass, which can be determined by measuring the attractive force exerted on it by another mass (M) a distance r away, viz:
gravitational mass = mgrav =
r2FMmGM
.
There is no mention of acceleration in this expression. Although these are two different concepts of mass, Newton asserted that, for the same object, these masses are identical, i.e., mgrav = miner. This is known as the
principle of equivalence.
Examples of gravitational and inertial mass:
Gravitational mass:
• related to the time it takes an object to fall from a
certain height.
• the weight of an object on a spring scale.
Inertial mass:
• the acceleration given to an object by a
compressed spring.
In fact, g is a vector and is often referred to as the
gravitational field, i.e., the region over which one mass
influences another. It is analogous to a magnetic field and
an electric field. Since is a vector, the resultant
gravitational field of a collection of mass is the vector sum
of the individual contributions.
(!g)
!g
Question 11.1: Three 5.0 kg masses are located at the
positions shown in the x-y plane. What is the resultant
force on the mass at caused by the other two masses? A
A
F1x = −Gm1mA
r12
= −(6.67 ×10−11N ⋅m2 /kg2)(5.0 kg)(5.0 kg)
(0.40 m)2
= −1.04 ×10−8N.
F2 = Gm2mA
r22
= (6.67 ×10−11N ⋅m2 /kg2) (5.0 kg)(5.0 kg)
(0.50 m)2
= 6.67 ×10−9 N.
∴F2x = −(6.67 ×10−9 N)cosθ = −0.80(6.67 ×10−9 N)= −(5.34 ×10−9 N),
and F2y = (6.67 ×10−9 N)sinθ = 0.60(6.67 ×10−9 N)= (4.00 ×10−9 N).
A
∴FRx = F1x + F2x
= (−1.04 ×10−8N) + (−5.34 ×10−9 N)
= −1.57 ×10−8N,
and FRy = F2y = 0.40 ×10−8N.
∴!FR = (−1.57 ×10−8N)i + (0.40 ×10−8N) j,
and !FR = 1.62 ×10−8N.
Question 11.2: Five objects, each of mass are
equally spaced on the arc of a semicircle of radius
An object of mass is located at
the center of curvature of the arc.
(a) What is the gravitational force acting on the
mass?
(b) What is the gravitational field at the center of
curvature of the arc?
M = 3.00 kg
R = 10.0 cm. m = 2.00 kg
2.00 kg
(a) The total force components acting on m are
and
Fx = GMm
R2+G
Mm
R2cos45! −G
Mm
R2cos45! −G
Mm
R2
= 0,
Fy = G Mm
R2sin45! +G Mm
R2+G Mm
R2sin45!
= G Mm
R21+ 2sin45!( ),
= (6.67 ×10−11 N ⋅m2 /kg2)(3.00 kg)(2.00 kg)
(0.10 m)2(2.414)
= 9.66 ×10−8 N.
∴!F = (9.66 ×10−8 N) j.
(b) The gravitational force acting on m is where
is the gravitational field. !F = m!g,
∴ !g =!Fm
= (9.66 ×10−8 N) j2.00 kg
= (4.83×10−8 m/s2) j.
!g Question 11.3: The Earth rotates about its N-S axis.
Assuming the Earth is a sphere, is the free-fall acceleration
measured at the equator the same as, greater than, or less
than the gravitational acceleration for a stationary Earth,
i.e., g = G
mE
RE2
?
View of the Earth looking at the
South pole and the FBD of
forces acting on an object
(O) at the equator; is the
normal force (equal to the
weight as measured on a scale) and is the gravitational
force. The net force is the centripetal force, i.e.,
where is the measured weight of the object. Then the
effective value of free-fall acceleration is
Since the Earth rotates though in 24 h,
Therefore, the effect is negligible.
!FN
!Fg = m!g
!a
FN
Fg
Fg − FN = mg − m ′g = mω2RE,
m ′g
′g = g −ω2RE, i.e., ′g < g.
2π
ω = 2π(24 × 60 × 60 s)
= 7.27 ×10−5 rad/s.
∴ω2RE = (7.27 ×10−5 rad/s)2(6.37 ×106 m) = 0.034 m/s2,
i.e., Δgg
= −0.0035.
O Gravitational potential energy
Above, we saw that the
acceleration of gravity at a
distance h above the Earth’s
surface was
g = G
mE
r2= G
mE
(h + RE )2.
Near the Earth’ surface, we showed that the
gravitational potential energy of an object of mass m at
a height h above the surface is
However, far from the Earth’s surface we must take into
account that In chapter 6, we defined a change
in potential energy as
where is a (conservative) force acting on an object
and is a general displacement.
In this case, for the gravitational force
Ug = mgh = mg(r − RE ).
g ∝ r−2.
dU = −!Fid!s,
!F
d!s
dU = −
!Fg id!r = Fg dr = G
mEm
r2dr.
∴U(r) = G
mEmr2∫ dr = GmEm
drr2∫ = −G
mEmr
+ U!,
where U! is an integration constant. We choose U! = 0, i.e., we define the gravitational potential energy to be zero at r = ∞.
We obtain a negative value for U(RE ) because we defined U(∞) = 0. In earlier chapters we usually defined
U(RE ) = 0. However, it does not cause difficulties as we normally work with changes in potential energy, i.e., ΔU, which is independent of our choice of U!.
r
U(r)
0
U(RE )
RE U(r) = −G
mEmr
←−G
mEmRE
= −mgRE
∴U(r) = G
mEm
r2dr∫ = GmEm dr
r2=∫ −G
mEmr
+ U!,
We have
U(r) = −GmEm
r= −G
mEmRE + h( ) = −G
mEm
RE 1 + hRE( ).
For small values of h, h
RE<< 1, then
U(r) = −G
mEmRE
1 +h
RE
⎛
⎝ ⎜
⎞
⎠ ⎟ −1
= −GmEmRE
1−h
RE
⎛
⎝ ⎜
⎞
⎠ ⎟
= −G
mEmRE
+ GmEmRE
2 h.
But G
mEmRE
= U(RE ) is constant and G
mERE
2 = g.
∴U(r) = U(RE ) + mgh ,
i.e., in moving an object a height h above the Earth’s surface increases the gravitational potential energy by
ΔU = mgh ,providing h << RE, a result we used in chapters 6 and 7.
Note that the weight of an object is ! F g = m! g , i.e.,
! g is a
vector and is called the gravitational field.
Question 11.4: A object is released from rest when it is
above the surface of a planet whose mass is
and radius Neglecting
air resistance,
(a) what is the speed of the object just before it strikes
the surface?
(b) Compare that speed with the speed it would have
achieved if the gravitational field were constant, i.e.,
the value at the surface.
4.0 ×106m
M = 4.0 ×1024kg R = 5.0 ×106m.
h
Ki = 0 : U(R + h)
Kf = 1
2mvf
2 : U(R)
(a) Using the conservation of
mechanical energy, i.e.,
we obtain Kf + Uf = Ki + Ui,
12
mvf2 −G Mm
R= 0 −G Mm
(R + h).
∴vf2 = 2(6.67 ×10−11N ⋅m2 /kg2)(4.0 ×1024kg)
× 1
5.0 ×106m− 1
9.0 ×106m
⎛⎝⎜
⎞⎠⎟
= 4.743×107(m/s)2.∴vf = 6.89 ×103 m/s.
(b) Assuming g is constant, g at the surface is
We have
g = G M
R2= (6.67 ×10−11N ⋅m2 /kg2) (4.0 ×1024kg)
(5.0 ×106m)2
= 10.7 m/s2.
ΔK = −ΔU. ∴ 1
2m ′v 2 = mgh,
i.e., ′v = 2gh = 2(10.7 m/s2)(4.0 ×106m)
= 9.25×103 m/s.
Question 11.5: When farthest from Earth, i.e., at apogee,
the Moon-Earth distance is 406,400 km. When closest to
Earth, i.e., at perigee, the Moon-Earth distance is
357,600 km. If the mass of the Earth is
what are the orbital speeds of the Moon at apogee and
perigee?
5.89 ×1024kg,
In the previous chapter we saw that providing there are no external forces/torques, then the angular momentum of a system like the Earth-Moon is conserved. The angular momentum of the Moon is
! L = ! r × m! v .
But at perigee and apogee ! r ⊥! v ,
so L = mvr .
∴mva ra = mvprp, i.e., va = vp
rpra
.
Also, mechanical energy is conserved, i.e.,
12
mva2 − G
MEmra
=12
mv p2 − G
MEmrp
.
∴va
2 − 2GMEra
= vp2 − 2G
MErp
.
Hence vp
2 rpra
⎛
⎝ ⎜
⎞
⎠ ⎟
2− 2G
MEra
= vp2 − 2G
MErp
,
i.e., vp
2 1−rp2
ra2
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ = 2GME
1rp
−1ra
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟ .
ra rp vp
va
Substituting for the given values
From earlier
vp2 0.2257( ) = 2.638×105 (m/s)2.
∴vp = 2.638×105 (m/s)2
0.2257= 1080 m/s.
va = vprpra
= (1080 m/s) (3.576 ×108m)
(4.064 ×108m)= 950 m/s.
Satellites
Consider a satellite, mass m, in a circular orbit, radius r, around the Earth (or a planet around the Sun). The gravitational force between the satellite and the Earth provides the centripetal acceleration.
∴G
mEmr2 =
mv2
r.
So, in a stable orbit
v2 =
GmEr
.
The time for one orbit T =
2πrv
i.e., v2 =
4π2r2
T2 = GmE
r= ′ g r,
where ′ g is the gravitational field acting on the satellite.
v
r m
ME
constant
∴T2 =
4π2
GmEr3 i.e., T2 ∝ r3.
~ Kepler’s 3rd Law of planetary motion ~ Note also
v = G
mEr
.
i.e., the velocity of a satellite in a stable orbit does not depend on its mass only the mass of the central object!
... well, it provides us a way of working out the mass of a planet. Take the Earth-Moon system for example ...
so what .. ?
Treating the Moon as a satellite around the Earth …
The speed of the Moon is
Then, from above
Note: we didn’t need to know the mass of the Moon, only
its distance from the Earth and its orbital period.
v = circumference of orbittime of orbit
= 2πrT
= 2π(3.84 ×108 m)(27.3× 24 × 60 × 60 s)
= 1.023 km/s.
v = G
mEr
mE = v2 rG
= (1.023×103 m/s)2(3.84 ×108 m)
6.67 ×10−11 N ⋅m2 /kg2
= 6.02 ×1024 kg.
Using a similar approach we can calculate the mass of the
Sun by considering the Earth as a satellite …
v = 2π(1.5×1011 m)(365× 24 × 60 × 60 s)
= 2.99 ×104 m/s.
∴mS = v2 rG
= (2.99 ×104 m/s)2 1.5×1011 m
6.67 ×10−11 N ⋅m2 /kg2
= 2.01×1030 kg.
Question 11.6: A 200 kg satellite circles the planet Roton
every 2.8 h in an orbit of radius If the radius
of Roton is what is
(a) the magnitude of the free-fall acceleration at the
surface of Roton, and
(b) the kinetic energy of the satellite?
1.20 ×107 m.
5.0 ×106m,
(a) The free-fall acceleration at the surface of a planet is
where M is its mass and R is its radius. However, we need
to find M. Using Kepler’s 3rd Law we have
where is the radius of the orbit and T is the orbital time.
Now
Hence
g = G M
R2,
T2 = 4π2
GMRs,
Rs
T = (2.8 h)(3600 s/h) = 1.008×104s,
∴M = 4π2
(6.67 ×10−11N ⋅m2 /kg2)
(1.20 ×107 m)3
(1.008×104s)2
= 1.01×1025kg.
g = (6.67 ×10−11N ⋅m2 /kg2) 1.01×1025kg
(5.0 ×106m)2
= 26.9 m/s2.
i.e., M = 4π2
GRs
3
T2
⎛
⎝⎜⎜
⎞
⎠⎟⎟
,
(b) For a satellite in orbit, the centripetal force
mv2
Rs= G Mm
Rs2
i.e., K = 12
mv2 = G Mm2Rs
= (6.67 ×10−11N ⋅m2 /kg2) (1.01×1025kg)(200 kg)
2(1.20 ×107 m)
= 5.61×109 J.
Question 11.7: Europa orbits Jupiter with a period of 3.55
days at an average distance of from Jupiter’s
center.
(a) Assuming the orbit is circular, what is the mass of
Jupiter?
(b) Another moon, Callisto, orbits Jupiter every 16.7
days. What is its average distance from Jupiter?
6.71×108 m
(a) Kepler’s 3rd Law states that
where and are the orbital period and orbital radius
of Europa, respectively, and is the mass of Jupiter.
(b) Also where subscript ca refers to
Callisto.
Teu
2 = 4π2
GMJReu
3 ,
Teu Reu
MJ
∴MJ =4π2Reu
3
GTeu2
= 4π2(6.71×108 m)3
(6.67 ×10−11 N ⋅m2 /kg2)(3.55× 24 × 60 × 60 s)
= 1.90 ×1027 kg.
Tca
2 = 4π2
GMJRca
3 ,
∴TcaTeu
⎛
⎝⎜⎞
⎠⎟
2
=RcaReu
⎛
⎝⎜⎞
⎠⎟
3
, i.e., Rca = ReuTcaTeu
⎛
⎝⎜⎞
⎠⎟
23
= (6.71×108 m) 16.7 d3.55 d
⎛⎝⎜
⎞⎠⎟
23
= 1.88×109 m.
Escape speed
The escape speed for a planet is defined as the minimum
initial speed of a projectile that enables it to escape the
influence of the planet’s gravitational field. Then
For the Earth
For an object launched from the Earth’s surface
• if the projectile will be “bound” and
will orbit the Earth in a circle or ellipse.
• if the projectile will be “unbound” and
will follow a hyperbolic path and leave the Earth and never
return.
Kf = 0 : U(∞) = 0
Ki =
12
mve2 : U(R)
Ki + Ui = Kf + Uf
i.e., 12
mve2 + U(R) = 0.
∴ 12
mve2 = G Mm
R,
i.e., ve = 2GMR
= 2gR.
ve = 2(9.81 m/s2)(6.37 ×106 m) = 11.2 km/s.
vi <11.2 km/s,
vi >11.2 km/s,
Question 11.8: Does it take more fuel, less fuel or the same
amount of fuel for a rocket to travel from the Earth to the
Moon as from the Moon to the Earth?
The work-kinetic energy theorem tells us that the
minimum work done to escape the Earth or Moon is the
same as the change in kinetic energy to achieve escape
speed.
But
The escape speed is Since and
then In fact,
Since the Earth’s escape speed is greater than the Moon’s
escape speed, it requires more work, i.e., more fuel, for a
spacecraft to travel from the Earth to the Moon than vice
versa.
∴W = ΔK = Ki − Kf .
Kf = 0,
∴W = 1
2mve
2.
ve = 2gR. RE > RM
gE > gM , ve(Earth) > ve(Moon).
ve(Earth) = 11.2 km/sve(Moon) = 2.36 km/s.
Question 11.9: Each planet in the solar system orbits at
its own particular speed around the Sun. If the orbital
speeds of the planets were doubled, which of the
following is true? Each planet would
A: assume a closer orbit around the Sun,
B: assume a more distant orbit around the Sun, or
C: escape the solar system.
D: Any or all of the above depending on the planet.
The escape speed from the Earth’s surface is
where are the mass and radius of the Earth,
respectively. So, the escape speed of a planet at a distance
R from the center of the Sun is
Earlier, we found that the orbital speed of a planet around the Sun is
So, if the orbital speed is doubled,
which is greater than the escape speed. Therefore, each
planet will escape the solar system (answer C). Note that
the orbital speed need only to be increased by a factor of
for a planet to escape the solar system.
ve =
2GMSR
.
ve =
2GMERE
,
ME and RE
v =
GMSR
.
′v ⇒ 2
GMSR
,
2
Question 11.10: A simple model of a black hole would be
to consider it a spherical object whose mass is so large that
the escape speed at the surface is greater than the speed of
light, c. The critical radius for the formation of a black
hole is called the Schwarzschild radius,
(a) Show that
where M is the mass of the black hole.
(b) Calculate the Schwarzschild radius for a black hole
whose mass is equal to 10 solar masses.
Take
RS.
RS = 2GM
c2,
c = 3×108 m/s.
(a) From earlier, we showed that the escape speed is
For a black hole,
(b)
ve = 2GM
R.
ve = c and R = RS,
i.e., c = 2GM
RS⇒ RS = 2GM
c2.
M = 10MSun = 10(2.0 ×1030 kg) = 2.0 ×1031 kg.
∴RS = 2(6.67 ×10−11 N ⋅m2 /kg2)(2.0 ×1031 kg)
(3×108 m/s)2
= 2.96 ×104 m (29.6 km).