NCERT Solutions for Class 11th: Ch 8 Gravitation Physics Science

Page No: 201

Excercises

8.1. Answer the following:
(a) You can shield a charge from electrical forces by putting
it inside a hollow conductor. Can you shield a body from the
gravitational influence of nearby matter by putting it inside a hollow
sphere or by some other means?

(b) An astronaut inside a small space ship orbiting around the
earth cannot detect gravity. If the space station orbiting around the
earth has a large size, can he hope to detect gravity?

(c) If you compare the gravitational force on the earth due to
the sun to that due to the moon, you would find that the Sun’s pull is
greater than the moon’s pull. (You can check this yourself using the
data available in the succeeding exercises). However, the tidal effect
of the moon’s pull is greater than the tidal effect of sun. Why?

Answer

(a) We cannot shiwld a body from the gravitational influence of nearby matter, because the gravitational force on a body due to nearby matter is independent of the presence of other matter, whereas it is not so in case of electric forces. It means the gravitational screens are not possible.

(b) Yes, if the size of the space station is large enough, then the astronaut will detect the change in Earth’s gravity (g).
(c) Tidal effect depends inversely upon the cube of the distance while, gravitational force depends inversely on the square of the distance. Since the distance between the Moon and the Earth is smaller than the distance between the Sun and the Earth, the tidal effect of the Moon’s pull is greater than the tidal effect of the Sun’s pull.

8.2. Choose the correct alternative:
(a) Acceleration due to gravity increases/decreases with increasing altitude.
(b) Acceleration due to gravity increases/decreases with increasing depth. (assume the earth to be a sphere of uniform density).
(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.
(d) The formula –G Mm(1/r2– 1/r1) is more/less accurate than the formula mg(r2r1) for the difference of potential energy between two points r2and r1distance away from the centre of the earth.

Answer

(a) decreases
(b) decreases
(c) mass of the body
(d) more.

8.3. Suppose
there existed a planet that went around the sun twice as fast as the
earth.What would be its orbital size as compaorange to that of the earth?

Answer

Time taken by the Earth
to complete one revolution around the Sun,

Te =
1 year
Orbital radius of the
Earth in its orbit, Re = 1 AU
Time taken by the
planet to complete one revolution around the Sun, TP = ½Te = ½ year
Orbital radius of the
planet = Rp

From Kepler’s
third law of planetary motion, we can write:
(Rp / Re)3 = (Tp / Te)2
(Rp / Re) = (Tp / Te)2/3
= (½ / 1)2/3  =  0.52/3  =  0.63
Hence, the orbital
radius of the planet will be 0.63 times smaller than that of the
Earth.

8.4. Io, one of the satellites of Jupiter, has an orbital period of
1.769 days and the radius of the orbit is 4.22 × 108
m. Show that the mass of Jupiter is about one-thousandth that of the
sun.

AnswerOrbital period of
I0 , TI0 = 1.769 days  =  1.769 × 24 × 60 × 60 s
Orbital radius of
I0 , RI0 = 4.22 × 108 m
Satellite I0 is
revolving around the Jupiter
Mass of the latter is given by the relation:
MJ = 4π2RI03 / GTI02    …..(i)
Where,
MJ = Mass of Jupiter
G = Universal gravitational constant
Orbital period of the earth,
Te = 365.25 days = 365.25 × 24 × 60 × 60 s
Orbital radius of the Earth,
Re = 1 AU = 1.496 × 1011 m
Mass of sun is given as:
Ms = 4π2Re3 / GTe      ……(ii)
Ms / MJ  =  (4π2Re3/ GTe2) × (GTI02/2RI03)   =   (Re3 × TI02) / (RI03 × Te2)
Substituting the values, we get:
= (1.769 × 24 × 60 × 60 / 365.25 × 24 × 60 × 60)2 × (1.496 × 1011 / 4.22 × 108)3
= 1045.04
Ms / MJ  ~ 1000
Ms ~ 1000 × MJ
Hence, it can be inferorange that the mass of Jupiter is about
one-thousandth that of the Sun.8.5. Let us assume that our galaxy consists of 2.5 × 1011
stars each of one solar mass. How long will a star at a distance of
50,000 ly from the galactic centre take to complete one revolution? Take
the diameter of the Milky Way to be 105 ly.
Answer

Mass of our galaxy
Milky Way, M = 2.5 × 1011 solar mass
Solar mass = Mass of
Sun = 2.0 × 1036
kg
Mass of our galaxy, M
= 2.5 × 1011 ×
2 × 1036 = 5
× 1041 kg
Diameter of Milky Way,
d = 105 ly
Radius of Milky Way, r
= 5 × 104 ly
1 ly = 9.46 ×
1015 m
r = 5 ×
104 × 9.46 ×
1015
=
4.73 ×1020 m

Since a star revolves
around the galactic centre of the Milky Way, its time period is given
by the relation:
T = ( 4π2r3/ GM)1/2
= [ (4 × 3.142 × 4.733 × 1060/ (6.67 × 10-11 × 5 × 1041) ]1/2
=   (39.48 × 105.82 × 1030/ 33.35 )1/2
= 1.12 × 1016 s
1 year = 365 × 324 × 60 × 60 s
1s = 1 / (365 × 324 × 60 × 60)  years
∴ 1.12 × 1016 s = 1.12 × 1016 / (365 × 24 × 60 × 60)  =  3.55 × 108 years.

8.6. Choose
the correct alternative:

(a) If
the zero of potential energy is at infinity, the total energy of an
orbiting satellite is negative of its kinetic/potential energy.

(b) The
energy requiorange to launch an orbiting satellite out of earth’s
gravitational influence is more/less than the energy requiorange to
project a stationary object at the same height (as the satellite) out
of earth’s influence.

Answer

(a) Kinetic energy
(b) Less

8.7. Does
the escape speed of a body from the earth depend on

(a)
the
mass of the body,

(b) the
location from where it is projected,

(c) the
direction of projection,

(d)
the
height of the location from where the body is launched?

Answer

The escape velocity is indpendent of the mass of the body and the direction of projection. It depends upon the gravitational potential at the point from where the body is launched. Since, this potential depends slightly on the latitude and height of the point, therefore, the escape velocity depends slightly on these factors.

8.8. A comet orbits the Sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.

Answer

A comet while going on elliptical orbit around the Sun has constant angular momentum and totaal energy at all locations but other quantities vary with locations.

8.9. Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem?

Answer

(a) Legs hold the entire mass of a body in standing position due to gravitational pull. In space, an astronaut feels weightlessness because of the absence of gravity. Therefore, swollen feet of an astronaut do not affect him/her in space.

(b) A swollen face is caused generally because of apparent weightlessness in space. Sense organs such as eyes, ears nose, and mouth constitute a person’s face. This symptom can affect an astronaut in space.

(c) Headaches are caused because of mental strain. It can affect the working of an astronaut in space.

(d) Space has different orientations. Therefore, orientational problem can affect an astronaut in space.

8.10. Choose the correct answer from among the given ones:The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig 8.12) (i) a, (ii) b, (iii) c, (iv) O.

Answer

Gravitational potential (V) is constant at all points in a spherical shell. Hence, the gravitational potential gradient (dV/dR) is zero everywhere inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, intensity is also zero at all points inside the spherical shell. This indicates that gravitational forces acting at a point in a spherical shell are symmetric.
If the upper half of a spherical shell is cut out (as shown in the given figure), then the net gravitational force acting on a particle located at centre O will be in the downward direction.

Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus, the gravitational intensity at centre O of the given hemispherical shell has the direction as indicated by arrow c.

8.11. Choose the correct answer from among the given ones:
For the problem 8.10, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.

Answer

Gravitational potential (V) is constant at all points in a spherical shell. Hence, the gravitational potential gradient (dV/dR) is zero everywhere inside the spherical shell. The gravitational potential gradient is equal to the negative of gravitational intensity. Hence, intensity is also zero at all points inside the spherical shell. This indicates that gravitational forces acting at a point in a spherical shell are symmetric.
If the upper half of a spherical shell is cut out (as shown in the given figure), then the net gravitational force acting on a particle at an arbitrary point P will be in the downward direction.

Since gravitational intensity at a point is defined as the gravitational force per unit mass at that point, it will also act in the downward direction. Thus, the gravitational intensity at an arbitrary point P of the hemispherical shell has the direction as indicated by arrow e.

Hence, the correct answer is (ii).
8.12. A
rocket is fiorange from the earth towards the sun. At what distance from
the earth’s centre is the gravitational force on the rocket
zero? Mass of the sun = 2 ×1030
kg, mass of the earth = 6 × 1024
kg. Neglect the effect of other planets etc. (orbital radius = 1.5
× 1011
m).
AnswerMass of the Sun, Ms
= 2 × 1030 kg

Mass of the Earth, Me
= 6 × 10 24 kg
Orbital radius, r
= 1.5 × 1011 m
Mass of the rocket = m

Let x be the
distance from the centre of the Earth where the gravitational force
acting on satellite P becomes zero.

From Newton’s law
of gravitation, we can equate gravitational forces acting on
satellite P under the influence of the Sun and the Earth as:
GmMs / (rx)2  =  GmMe / x2
[ (rx) / x ]2  =  Ms / Me
(rx) / x  =  [ 2 × 1030 / 60 × 1024]1/2  =  577.35
1.5 × 1011x = 577.35x
578.35x = 1.5 × 1011
x = 1.5 × 1011 / 578.35  =  2.59 × 108 m.

8.13.How
will you ‘weigh the sun’, that is estimate its mass? The
mean orbital radius of the earth around the sun is 1.5 × 108
km.

Answer

Orbital radius of the Earth around the Sun, r = 1.5 × 1011 m
Time taken by the Earth to complete one revolution around the Sun,
T = 1 year = 365.25 days
= 365.25 × 24 × 60 × 60 s
Universal gravitational constant, G = 6.67 × 10–11 Nm2 kg–2
Thus, mass of the Sun can be calculated using the relation,
M = 4π2r3 / GT2
= 4 × 3.142 × (1.5 × 1011)3/ [ 6.67 × 10-11 × (365.25 × 24 × 60 × 60)2]
= 2 × 1030 kg
Hence, the mass of the Sun is 2 × 1030 kg.

8.14. A
Saturn year is 29.5 times the earth year. How far is the Saturn from
the sun if the earth is 1.50 ×108
km away from the sun?

Answer

Distance of the Earth
from the Sun, re = 1.5 ×
108 km = 1.5 ×
1011 m
Time period of the
Earth = e
Time period of Saturn,
Ts = 29. 5 Te
Distance of Saturn from
the Sun = rs

From Kepler’s
third law of planetary motion, we have
T = (4π2r3 / GM)1/2
For Saturn and Sun, we
can write
rs3 / re3  =  Ts2 / Te2
rs = re(Ts / Te)2/3
= 1.5 × 1011 (29.5 Te / Te)2/3
= 1.5 × 1011 (29.5)2/3
= 14.32 × 1011 m
Hence, the distance
between Saturn and the Sun is 1.43 × 1012 m.

8.15. A
body weighs 63 N on the surface of the earth. What is the
gravitational force on it due to the earth at a height equal to half
the radius of the earth?

Answer

Weight of the body, W
= 63 N
Acceleration due to
gravity at height h from the Earth’s surface is given by
the relation:
g‘ = g / [1 + ( h / Re) ]2
Where,
g = Acceleration due to
gravity on the Earth’s surface

Re = Radius of
the Earth
For h = Re / 2
g‘ = g / [(1 + (Re / 2Re) ]2
= g / [1 + (1/2) ]2  =  (4/9)g
Weight of a body of
mass m at height h is given as:
W‘ = mg
= m × (4/9)g  =  (4/9)mg
= (4/9)W
= (4/9) × 63  =  28 N.

8.16. Assuming
the earth to be a sphere of uniform mass density, how much would a
body weigh half way down to the centre of the earth if it weighed 250
N on the surface?

Answer

Weight of a body of
mass m at the Earth’s surface, W = mg =
250 N
Body of mass m is located at depth, d = (1/2)Re
Where,
Re = Radius of the Earth

Acceleration due to
gravity at depth g (d) is given by the relation:

g‘ = (1 – (d / Re)g
= [1 – (Re / 2Re]g  =  (1/2)g
Weight of the body at
depth d,

W‘ = mg
= m × (1/2)g  =  (1/2) mg  =  (1/2)W
= (1/2) × 250  =  125 N

8.17. A
rocket is fiorange vertically with a speed of 5 km s–1
from the earth’s surface. How far from the earth does the
rocket go before returning to the earth? Mass of the earth = 6.0 ×
1024
kg; mean radius of the earth = 6.4 ×
106
m; G=
6.67 ×
10–11
N m2
kg2.


Answer

Velocity of the rocket,
v = 5 km/s = 5 × 103
m/s
Mass of the Earth, Me = 6 × 1024 kg
Radius of the Earth, Re = 6.4 × 106 m
Height reached by
rocket mass, m = h
At the surface of the
Earth,
Total energy of the
rocket = Kinetic energy + Potential energy
= (1/2)mv2 + (-GMem / Re)
At highest point h,
v = 0
And, Potential energy = -GMem / (Re + h)
Total energy of the
rocket = 0 + [ -GMem / (Re + h) ]
= -GMem / (Re + h)
From the law of
conservation of energy, we have
Total energy of the
rocket at the Earth’s surface = Total energy at height h

(1/2)mv2 + (-GMem / Re) = -GMem / (Re + h)
(1/2)v2 = GMe [ (1/Re) – 1 / (Re + h) ]
= GMe[ (Re + hRe)  / Re(Re+ h) ]
(1/2)v2 = gReh / (Re + h)
Where g = GM / Re2 = 9.8 ms-2
v2 (Re + h) = 2gReh
v2Re = h(2gRev2)
h = Rev2 / (2gRe –  v2)
= 6.4 × 106 × (5 × 103)2 / [ 2 × 9.8 × 6.4 × 106 – (5 × 103)2
h = 1.6 × 106 m
Height achieved by the
rocket with respect to the centre of the Earth = Re + h
= 6.4 × 106 + 1.6 × 106  =  8 × 106 m.

8.18. The escape speed of a
projectile on the earth’s surface is 11.2 km s–1.
A body is projected out with thrice this speed. What is the speed of
the body far away from the earth? Ignore the presence of the sun and
other planets.

Answer

Escape velocity of a
projectile from the Earth, vesc = 11.2 km/s
Projection velocity of
the projectile, vp = 3vesc
Mass of the projectile
= m
Velocity of the
projectile far away from the Earth = vf
Total energy of the
projectile on the Earth = (1/2)mvp2 – (1/2)mvesc2
Gravitational potential
energy of the projectile far away from the Earth is zero.
Total energy of the
projectile far away from the Earth = (1/2)mvf2
From the law of
conservation of energy, we have
(1/2)mvp2 – (1/2)mvesc2  =  (1/2)mvf2
vf = ( vp2vesc2 )1/2
= [ (3vesc)2vesc2]1/2
= √8 vesc
= √8 × 11.2  =  31.68 km/s.



8.19. A
satellite orbits the earth at a height of 400 km above the surface. How
much energy must be expended to rocket the satellite out of the earth’s
gravitational influence? Mass of the satellite = 200 kg; mass of the
earth = 6.0 ×1024 kg; radius of the earth = 6.4 ×106 m; G = 6.67 × 10–11 N m2 kg–2.


Answer

Mass of the Earth, M
= 6.0 × 1024 kg
Mass of the satellite,
m = 200 kg
Radius of the Earth, Re
= 6.4 × 106 m
Universal gravitational
constant, G = 6.67 × 10–11 Nm2kg–2
Height of the
satellite, h = 400 km = 4 × 105 m = 0.4 ×106
m
Total energy of the
satellite at height h = (1/2)mv2 + [ -GMem / (Re + h) ]
Orbital velocity of the
satellite, v = [ GMe / (Re + h) ]1/2
Total energy of height,
h = (1/2)GMem / (Re + h) – GMem / (Re + h)  =  -(1/2)GMem / (Re + h)
The negative sign
indicates that the satellite is bound to the Earth. This is called
bound energy of the satellite.
Energy requiorange to send
the satellite out of its orbit = – (Bound energy)
= (1/2) GMem / (Re + h)
= (1/2) × 6.67 × 10-11 × 6 × 1024 × 200 / (6.4 × 106 + 0.4 × 106)
= 5.9 × 109 J.

8.20. Two stars each of one
solar mass (= 2× 1030 kg) are approaching each other
for a head on collision. When they are a distance 109 km, their
speeds are negligible. What is the speed with which they collide? The
radius of each star is 104 km. Assume the stars to remain undistorted
until they collide. (Use the known value of G).

Answer

Mass of each star, M = 2 × 1030 kg
Radius of each star, R = 104 km = 107 m
Distance between the stars, r = 109 km = 1012m
For negligible speeds, v = 0 total energy of two stars separated at distance r
= [ -GMM / r ] + (1/2)mv2
[ -GMM / r ] + 0  ….(i)
Now, consider the case when the stars are about to collide:
Velocity of the stars = v
Distance between the centers of the stars = 2R
Total kinetic energy of both stars = (1/2) Mv2 + (1/2)Mv2 = Mv2
Total potential energy of both stars = -GMM / 2R
Total energy of the two stars = Mv2 – GMM / 2R    ….(ii)
Using the law of conservation of energy, we can write:
Mv2 – GMM / 2R =  -GMM / r
v2 = -GM / r + GM / 2R
= GM [ (-1/r) + (1/2R) ]
= 6.67 × 10-11 × 2 × 1030[ (-1/1012 ) + (1 / 2 × 107) ]
~ 6.67 × 1012
v = ( 6.67 × 1012)1/2  =  2.58 × 106 m/s.


8.21. Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a
horizontal table. What is the gravitational force and potential at the mid point of the line joining the centers of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium
stable or unstable?

Answer

Grvitational field at the mid-point of the line joining the centres of the two spheres
= GM/(r/2)2 (alog negative r) + GM/(r/2) (along r) = 0

Gravitational potential at the midpoint f the line joining the centres of the two spheres is

V = – GM/r/2 + (-GM/r/2) = -4GM/r = -4 × 6.67 × 10-11 × 100/1.0
= -2.7 × 10-8 J/Kg

As the effective force on the body placed at mid-point is zero, sso the body is in equilibrium. If the body is displaced a little towards either mass body from its equilibrium position, it will not return back to its inital position of equilibrium. Hence, the body is in unstable equilibrium.

8.22. As you have learnt in
the text, a geostationary satellite orbits the earth at a height of
nearly 36,000 km from the surface of the earth. What is the potential
due to earth’s gravity at the site of this satellite? (Take the
potential energy at infinity to be zero). Mass of the earth = 6.0
× 1024
kg, radius = 6400 km.

Answer

Mass of the Earth, M
= 6.0 × 1024 kg
Radius of the Earth, R
= 6400 km = 6.4 × 106 m
Height of a
geostationary satellite from the surface of the Earth,

h = 36000 km =
3.6 × 107 m
Gravitational potential
energy due to Earth’s gravity at height h,

= -GM / (R + h)
= – 6.67 × 10-11 × 6 × 1024 / (3.6 × 107 + 0.64 × 107)
= -9.4 × 106 J/kg.

8.23. A star 2.5 times the
mass of the sun and collapsed to a size of 12 km rotates with a speed
of 1.2 rev. per second. (Extremely compact stars of this kind are
known as neutron stars. Certain stellar objects called pulsars belong
to this category). Will an object placed on its equator remain stuck
to its surface due to gravity? (Mass of the sun = 2 × 1030
kg).

Answer

A body gets stuck to
the surface of a star if the inward gravitational force is greater
than the outward centrifugal force caused by the rotation of the
star.
Gravitational force, fg = – GMm / R2
Where,
M = Mass of the
star = 2.5 × 2 × 1030 = 5 × 1030 kg
m = Mass of the
body
R = Radius of
the star = 12 km = 1.2 ×104 m
fg = 6.67 × 10-11 × 5 × 1030 × m / (1.2 × 104)2  =  2.31 × 1011 m N
Centrifugal force, fc
= mrω2
ω
= Angular speed = 2πν

ν = Angular
frequency = 1.2 rev s–1
fc =
mR (2πν)2

= m ×
(1.2 ×104) × 4 × (3.14)2 ×
(1.2)2 = 1.7 ×105m N
Since fg
> fc, the body will remain stuck to the surface
of the star.

8.24. A
spaceship is stationed on Mars. How much energy must be expended on the
spaceship to launch it out of the solar system? Mass of the space ship =
1000 kg; mass of the Sun = 2 × 1030 kg; mass of mars = 6.4 × 1023 kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 × 108kg; G= 6.67 × 10–11 m2kg–2.

Answer

Mass
of the spaceship, ms
= 1000 kg
Mass
of the Sun, M
= 2 × 1030
kg
Mass
of Mars, mm
= 6.4 × 10 23
kg
Orbital
radius of Mars, R
= 2.28 × 108 kg
=2.28 × 1011m
Radius
of Mars, r =
3395 km = 3.395 × 106
m
Universal
gravitational constant, G = 6.67 × 10–11
m2kg–2
Potential
energy of the spaceship due to the gravitational attraction of the
Sun = -GMms / R
Potential
energy of the spaceship due to the gravitational attraction of Mars = -GMmms / r
Since
the spaceship is stationed on Mars, its velocity and hence, its
kinetic energy will be zero.
Total
energy of the spaceship =  -GMms/ R
GMmms / r
= -Gms[ (M / R) + (mm / r) ]
The negative sign
indicates that the system is in bound state.
Energy requiorange for
launching the spaceship out of the solar system
= – (Total energy
of the spaceship)
= Gms[ (M / R) + (mm / r) ]
= 6.67 × 10-11 × 103 × [ (2 × 1030/ 2.28 × 1011) + (6.4 × 1023/ 3.395 × 106 ) ]
= 596.97 × 109  =  6 × 1011 J.

8.25. A rocket is fiorange
‘vertically’ from the surface of mars with a speed of 2
km s–1. If 20% of its initial energy is lost due to Martian
atmospheric resistance, how far will the rocket go from the surface
of mars before returning to it? Mass of mars = 6.4× 1023 kg;
radius of mars = 3395 km; G = 6.67× 10-11 N m2
kg–2.

Answer

Initial velocity of the
rocket, v = 2 km/s = 2 × 103 m/s
Mass of Mars, M
= 6.4 × 1023 kg
Radius of Mars, R
= 3395 km = 3.395 × 106 m
Universal gravitational
constant, G = 6.67× 10–11 N m2 kg–2
Mass of the rocket = m
Initial kinetic energy
of the rocket = (1/2)mv2
Initial potential
energy of the rocket = -GMm / R
Total
initial energy = (1/2)mv2– GMm / R
If
20 % of initial kinetic energy is lost due to Martian atmospheric
resistance, then only 80 % of its kinetic energy helps in reaching a
height.
Total
initial energy available = (80/100) × (1/2) mv2 – GMm / R  =  0.4mv2 – GMm / R
Maximum
height reached by the rocket = h
At
this height, the velocity and hence, the kinetic energy of the rocket
will become zero.
Total
energy of the rocket at height h = -GMm / (R + h)

Applying
the law of conservation of energy for the rocket, we can write:

0.4mv2 – GMm / R  =  -GMm / (R + h)
0.4v2 = GM / R  –  GM / (R + h)
= GMh / R(R + h)
(R + h) / h  =  GM / 0.4v2R
R / h  =  ( GM / 0.4v2R )  –  1
h = R / [ (GM / 0.4v2R) – 1 ]
= 0.4R2v2/ (GM – 0.4v2R)
= 0.4 × (3.395 × 106)2 × (2 × 103)2/ [ 6.67 × 10-11 × 6.4 × 1023  –  0.4 × (2 × 103)2 × (3.395 × 106) ]
= 18.442 × 1018 / [ 42.688 × 1012  –  5.432 × 1012 ]
= 18.442 × 106 / 37.256
= 495 × 103 m = 495 km.