Mod-10

Module 10: Gravitation Kepler’s Law: In the early 1600's, German mathematician and astronomer Johannes Kepler mathematically analyzed known astronomical data in order to develop three laws to describe the motion of planets about the sun. Kepler's three laws emerged from the analysis of data carefully collected over a span of several years by his Danish predecessor and teacher, Tycho Brahe. Kepler's three laws of planetary motion can be briefly described as follows: •

The path of the planets about the sun are elliptical in shape, with the center of the sun being located at one focus. (The Law of Ellipses)

•

An imaginary line drawn from the center of the sun to the center of the planet will sweep out equal areas in equal intervals of time (

•

dA = cons tan t ). (The Law of Equal Areas) dt

The square of the period of any planet is proportional to the cube of the planet's mean distance from the sun (T 2 = CR 3 ). (The Law of Harmonies)

Period (s)

Planet

T2/R3 (s2/m3)

Average Dist. (m)

Newton’s Law: Isaac Newton compared the acceleration of the moon to the acceleration of objects on earth. Believing that gravitational forces were responsible for each, Newton was able to draw an important conclusion about the dependence of gravity upon distance. This comparison led him to conclude that “ALL objects attract each other with a force of gravitational attraction. This force of gravitational attraction is directly dependent upon the masses of both objects and inversely proportional to the square of the distance which separates their centers”. Newton's conclusion about the magnitude of gravitational forces is summarized symbolically as:

Fgrav ∝

m1 ∗ m 2 d2

- - - - - - - - - - - - - - - - (1)

Where, Fgrav represents the force of gravity between two objects m1 represents the mass of object 1 m2 represents the mass of object 2 d represents the distance separating the object’s centers. Since the gravitational force is directly proportional to the mass of both interacting objects, more massive objects will attract each other with a greater gravitational force. So, as the mass of either object increases, the force of gravitational attraction between them also increases.

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Since gravitational force is inversely proportional to the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases. The proportionalities expressed by Newton's universal law of gravitation is represented graphically by the following illustration. Observe how the force of gravity is directly proportional to the product of the two masses and inversely proportional to the square of the distance of separation.

Another means of representing the proportionalities is to express the relationships in the form of an equation using a constant of proportionality. This equation is shown below.

Fgrav =

Gm1m 2 d2

− − − − − − − ( 2)

Where, G represents the universal gravitational constant (G = 6.67x10-11 N-m2/kg2)

Gravitation and the Principle of Superposition: For a group of particles, the net gravitational force exerted on any one of them can be found by principle of superposition, which states that “The total force on a point particle is equal to the sum of all the forces on the particle” For n interacting particles, the principle of superposition for gravitational force is written as,

     F1 = F12 + F13 + F14 _ ....... + F1n n  = ∑ F1i i=2



Where, F1i : force of the ith particle on particle 1. For a real object with a continuous distribution of particles:

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F1 = ∫ dF1 dF1 = G

m1dm

r2 m F1 = ∫ G 21 dm r

Gravitational force from a thin ring: dm

F X = ∫ dF X = ∫G

m1dm 2

r m dm x = ∫G 12 r r m xdm = ∫G 1 3 r m1 xdm = ∫G 3

(x

=G

2

r

cos θ

+ R2

)

θ x

dF

dFY dFX

m1

R

2

m1mx

(x

2

+ R2

)

3 2

• The force points toward the ring. • Fx = 0 if x = 0

Other Newton's results: • A uniform spherical shell of matter attracts a particle outside as if all the shell mass was concentrated at the center. • Similarly for a sphere of matter. • Like the case of a ring, a particle inside a spherical shell of matter feels zero gravitational force from the shell.

The Value of acceleration of gravity (g): To determine the force of gravity (Fgrav) with which an object of mass m was attracted to the earth an equation was given by Newton Fgrav = mg Now in this unit, a second equation has been introduced for calculating the force of gravity with which an object is attracted to the earth.

Fgrav =

Gm1m 2 d2

Where, d represents the distance from the center of the object to the center of the earth. In the first equation above, g is referred to as the acceleration of gravity. It's value is 9.8 m/s2 on Earth. First, set both expressions for the force of gravity are set equal to each other.

mg =

GM earth m d2

Now observe that the mass of the object - m - is present on both sides of the equal sign. Thus, m can be canceled from the equation. This leaves us with an equation for the acceleration of gravity.

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g=

GM earth d2

The above equation demonstrates that the acceleration of gravity is dependent upon the mass of the earth (approx. 5.98x1024 kg) and the distance (d) that an object is from the center of the earth. The table below shows the value of g at various locations from Earth's center. Location

Distance from Earth's center (m)

Value of g m/s2

Earth's surface

6.38 x 106 m

9.8

1000 km above surface 2000 km above surface 4000 km above surface

6

7.33

6

5.68

7

3.70

7

7.38 x 10 m 8.38 x 10 m 1.04 x 10 m

6000 km above surface

1.24 x 10 m

2.60

9000 km above surface

1.54 x 107 m

1.69

50000 km above surface

7

5.64 x 10 m

0.13

As is evident from both the equation and the table above, the value of g varies inversely with the distance from the center of the earth. In fact, the variation in g with distance follows an inverse square law where g is inversely proportional to the distance from earth's center. This inverse square relationship means that as the distance is doubled, the value of g decreases by a factor of 4; as the distance is tripled, the value of g decreases by a factor of 9; and so on. This inverse square relationship is depicted in the graphic at the right.

Variation of g with Altitude: Let P be a point on the surface of the earth and Q another point at an altitude h. R is the radius of the earth and M is the mass of the earth. A body of mass m when is placed at the point P experience a force mg towards the centre of the earth. ∴

mg =

GMm - - - - - - - - - - - - - - - - - (1) R2

Q

When the body is at Q, let the acceleration due to gravity be g’,

mg ′ =

GMm (R + h ) 2

P - - - - - - - - - - - - - - - - - (2)

R

Dividing Eq. (2) by Eq. (1),

mg ′ R2  R  = =  2 mg (R + h ) R +h g′  R + h  =  g  R 

−2

h  = 1 +  R  

Earth

2

−2

= 1−

2h R

4

h

 2h  g ′ = g 1 − − − − − − − − − − − − (3 ) R  

∴

Therefore, the acceleration due to gravity (g) decreases with altitude.

Variation of g with Depth: Let g and g’ be the acceleration due to gravity at P and Q respectively. At P, the whole mass of the earth attracts the body and at Q it is attracted by the mass of the earth of radius (R-h).

mg =

GMm - - - - - - - - - - - - - - - - - (1) R2

GM ′m mg ′ = (R − h ) 2

h

P Q

- - - - - - - - - - - - - - - - - (2)

R

If ρ is the mean density of the earth then the mass of the earth becomes, M =

4 4 πR 3 ρ and M ′ = π (R − h )3 ρ 3 3

Dividing Eq. (2) by Eq. (1) and substituting the value of M and M’ we get,

4 π (R − h ) 3 ρ mg ′ M′ R2 R2 3 = x = x 4 mg (R − h ) 2 M (R − h ) 2 πR 3 ρ 3 g′ R − h h = = 1− g R R h  ∴ g ′ = g 1 −   R Therefore, the acceleration due to gravity (g) decreases with depth.

Variation of g with Rotation of the Earth:

B

Consider a body of mass m on the surface of the earth at latitude ϕ,

ϕ O

Here,

P

ϕ

Q

A

OP = R BP = Rcosϕ

Let ω be the angular velocity of the earth. Force mg acts along PO. Let us resolve mg into two components mg cosϕ along PB and mg sinϕ along PA.

Due to the rotation of the earth, a body at P experiences an outward force,

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mv 2 = mrω 2 ; r The net force along PB = mg cosϕ - mRω2cosϕ The resultant force experienced by P is along PQ which is the resultant of the force along PA and PB.

F= ∴

( mg sin ϕ ) 2 + (mg cos ϕ − mRω 2 cos ϕ )

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mg ′ = m 2 g 2 sin 2 ϕ + m 2 g 2 cos 2 ϕ − 2m 2 gRω 2 cos 2 ϕ + m 2 R 2ω 4 cos 2 ϕ = mg sin 2 ϕ + cos 2 ϕ −

Now we know that

2Rω 2 cos 2 ϕ R 2ω 4 cos 2 ϕ + g g2

Rω 2 1 = g 289

[R = 6.4x108 cm, ω = 2πn, n =

1 ] 24 x 60 x 60

Therefore, last term can be neglected, 1

∴

 2Rω 2 cos 2 ϕ  2  mg ′ = mg 1 −  g   2 2  Rω cos ϕ   g ′ = g 1 −  g  

(i)

At the equator ϕ = 0o, cos ϕ = 1

∴

(ii)

 Rω 2 g ′ = g 1 − g 

   

At the poles ϕ = 90o, cosϕ = 0 ∴

g’ = g.

Therefore, acceleration due to gravity at the poles is greater than that at the equator.

Example: Suppose a tunnel could be dug through the earth from one side to the another along a diameter as shown in Figure. Show that the motion of a particle dropped into the tunnel is simple harmonic motion. Neglect all frictional force and assume that earth has uniform density. The gravitational attraction of the earth for the particle at a distance R from the of earth arises entirely from that portion of matter of the earth in shells internal to position of that particle. The external mass has no force on the particle. Let us consider the earth has uniform density ρ. Then the mass inside a sphere of r is,

M ′ = ρv ′ = ρ

4 3 πr 3

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r m

centre the radius

This mass can be treated as, it were concentrated at the centre of the earth for gravitational purpose. Hence the force on the particle of mass m is

F =−

GM ′m r2

The minus sign indicates that force is attractive.

Hence,

∴

F = -G

∴

F = -kr

4πr 3 ρm 4πm   = -  Gρ r 2 3  3r 

Gρ 4πm is a constant which is denoted as k. The force is therefore proportional to the displacement r 3

but oppositely directed. This is exactly the criterion for simple harmonic motion. The motion of the particle is simple harmonic.

Escape velocity: Escape velocity is defined as the velocity with which a body has to be projected vertically upwards from the earth’s surface so that it escape the earth’s gravitational field altogether. If v is the escape velocity, then the initial kinetic energy of projection

1 2 mv must be equal to the work done in 2

moving the body from the surface of the earth to infinity. In such case, the body does not return to the earth’s surface. Let mass of the earth and that of the body be M and m radius of the earth is R. Suppose a body is at a distance x from the earth’s centre. The force of gravity acting on the body =

GMm x2

If the body is to be moved through a small distance dx away from the earth surface, the work done,

 GMm  dw =  dx 2   X  Work done in moving the body from the surface of the earth to infinity ∞

∞

GMm 1 GMm W = ∫ dw = ∫ dx = GMm − = 2 xR R R x Therefore,

1 GMm mv 2 = 2 R v=

[K.E. = Work done]

2GM R

Also at the surface of the earth,

∴ ∴

v=

GMm R2 GM = gR 2

mg =

2gR 2 = 2gR R

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Planetary and Satellite Motion: Satellite is any object which is orbiting the earth, sun or other massive body. Satellites can be categorized as natural satellites or man-made satellites. The moon, the planets and comets are examples of natural satellites. Accompanying the orbit of natural satellites are a host of satellites launched from earth for purposes of communication, scientific research, weather forecasting, intelligence, etc. Whether a moon, a planet, or some man-made satellite, every satellite's motion is governed by the same physics principles and described by the same mathematical equations. The motion of an orbiting satellite can be described by the same motion characteristics as any object in circular motion. The velocity of the satellite would be directed tangent to the circle at every point along its path. The acceleration of the satellite would be directed towards the center of the circle - towards the central body which it is orbiting. And this acceleration is caused by a net force which is directed inwards in the same direction as the acceleration.

This centripetal force is supplied by gravity - the force which universally acts at a distance between any two objects which have mass. Were it not for this force, the satellite in motion would continue in motion at the same speed and in the same direction.

Mathematics of Satellite Motion: The motions of objects are governed by Newton's laws. The same simple laws which govern the motion of objects on earth also extend to the heavens to govern the motion of planets, moons, and other satellites. The mathematics which describes a satellite's motion are the same mathematics presented for circular motion. Consider a satellite with mass Msat orbiting a central body with a mass of mass MCentral. The central body could be a planet, the sun or some other large mass capable of causing sufficient acceleration on a less massive nearby object. If the satellite moves in circular motion, then the net centripetal force acting upon this orbiting satellite is given by the relationship

Fnet =

M satV 2 R

This net centripetal force is the result of the gravitational force which attracts the satellite towards the central body, and can be represented as

Fgrav =

GM sat MCentral R2

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Since Fgrav = Fnet, the above expressions for centripetal force and gravitational force are equal. Thus,

M satV 2 GM sat M Central = R R2 Observe that the mass of the satellite is present on both sides of the equation; thus it can be canceled by dividing through by Msat. Then both sides of the equation can be multiplied by R, leaving the following equation.

V2 =

GM Central R

Taking the square root of each side, leaves the following equation for the velocity of a satellite moving about a central body in circular motion

V =

GM Central R

where G = 6.67 x 10-11 N m2/kg2, Mcentral = the mass of the central body about which the satellite orbits, and R = the radius of orbit for the satellite. Similar reasoning can be used to determine an equation for the acceleration of our satellite that is expressed in terms of masses and radius of orbit. The acceleration value of a satellite is equal to the acceleration of gravity of the satellite at whatever location which it is orbiting. The equation for the acceleration of gravity is

g=

GM central R2

Thus, the acceleration of a satellite in circular motion about some central body is given by the following equation

a=

GM central R2

where G = 6.67 x 10-11 N m2/kg2, Mcentral = the mass of the central body about which the satellite orbits, and R = the average radius of orbit for the satellite. The period of a satellite (T) and the mean distance from the central body (R) are related by the following equation:

T2 4π 2 = R 3 GM central where T = the period of the satellite, R = the average radius of orbit for the satellite (distance from center of central planet), and G = 6.67 x 10-11 N m2/kg2. There is an important concept evident in all three of these equations - the period, speed and the acceleration of an orbiting satellite are not dependent upon the mass of the satellite.

V =

GM Central R

a=

T2 4π 2 = R 3 GM central

GM central R2

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None of these three equations has the variable Msatellite in them. The period, speed and acceleration of a satellite is only dependent upon the radius of orbit and the mass of the central body which the satellite is orbiting. Just in the case of the motion of projectiles on earth, the mass of the projectile has no effect upon the acceleration towards the earth and the speed at any instant. Whenever air resistance is negligible and all forces but gravity are nonexistent, the mass of the moving object becomes a non-factor. Such is the case of orbiting satellites.

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SOLVED EXAMPLES Example 1: Suppose a tunnel could be dug through the earth from one side to the other along a diameter. If mail were delivered through this chute, how much time would elapse between deposit at one end and delivery at the other end? Solution: The mail would execute simple harmonic motion and the period of this motion is

T = 2π

Let us take

m 3m = 2π = k Gρ 4 πm

3π Gρ

ρ = 5.51 x 103 kg/m3 and G = 6.67 x 10-11 N.m2 / kg2. This gives

T=

3π = 5050 s =84.2 min Gρ

The time of delivery is one-half period, or about 42 minutes. Note that this time is independent of the mass of the mail. Example 2: The fact that g varies from place to place over the earth’s surface drew attention when Jean Richer in 1672 took a pendulum clock from Paris to Cayenne, French Guiana, and found that it lost 2.5 min/day. If g = 9.81 m/s2 in Paris, what is g in Cayenne? Solution: We know that the period

T1 ∝ 1/ g1 Then we can write

T1 = T2

g2 , or g 2 = g1 [ T1 / T2 ]2 g1

Therefore, Value of g at Cayenne =

g 2 = [9.81 m / s 2 ][86400 sec/(86400 + 150) sec] 2 = 9.776 m / s 2

Example 3: The mean distance of Mars from the sun is 1.52 times that of Earth from the sun. Find the number of years required for Mars to make one revolution about the sun. Solution: Since square of period of a orbit is proportional to the cube of the distance i.e.

Tmars r = [ mars ]3 / 2 , or Tearth rearth

Tmars = [ Tearth ][

T 2 ∝ r 3 , we can write

rmars 3 / 2 ] = [1 yr ][1.52 ]3 / 2 = 1.87 rearth

yr

Example 4: Determine the mass of the earth from the period T and the radius R of the moon’s orbit about the earth: T = 27.3 days and r = 3.845 x 105 km. Solution: We can calculate the mass of the earth from

4 π 2r 3 [ 4 π 2 ][3.845 x10 8 m]3 M= = = 6.04 x10 24 kg. 2 2 2 6 2 GT [6.67 x 10 - 11 N.m / kg ][2.359 x10 s ] 11

Example 5: How far from Earth must a space probe be along a line towards the sun so that the sun’s gravitational pull on the probe balances Earth’s pull? Solution: At the point where the forces balance

GM e m GM s m = , ………….. [a] r12 r22 where Me = the mass of the earth, Ms = the mass of the sun, m = the mass of the space probe, r 1 = the distance of the probe from the centre of the earth, r2 = the distance of the space probe from the centre of the sun. If d is the distance from the centre of the earth to the centre of the sun, then we can write r2 = d – r1. Using this result, we can write Eq. [a]

Me Ms = , ……………….[b] r12 [d − r1 ]2 After a little algebra, we can write

r1 =

d Me Ms + Me

=

[150 x10 9 m] 5.98 x10 24 kg 1.99 x10

Example 6: A rocket is accelerated to speed

30

kg + 5.98x10

24

kg

= 2.6x10 8 m

v = 2 gR e near earth’s surface where earth’s radius is Re, and

it then coasts upward. [a] Show that it will escape from earth. [b] Show that very far from earth its speed will be

v = 2gR e . Solution: [a] Initially the rocket is at earth’s surface and the potential energy is

Ui = − GMm / R e = −GMmR e / R 2e = −mgR e where we have used the relationship

g = GM / R 2e .

The initial kinetic energy is

Ki = where we have used

1 mv 2 = 2mgR e 2

v = 2 gR e .

If the rocket can escape then conservation of energy must lead to a positive kinetic energy no matter how far from earth it gets. Take the final potential energy to be zero and let Kf be the final kinetic energy. Then,

Ui + K i = U f + K f leads to K f = Ui + K i = −mgR e + 2mgR e = mgR e The result is positive and the rocket has enough kinetic energy to escape the gravitational pull of earth. [b] Now

Kf =

Therefore,

1 mv 2f = mgR e . 2

v f = 2gR e 12

Problems: Gravitation 1. How far from the earth must a body be along a line toward the sun so that the sun’s gravitational pull balances the earth’s? The sun is 9.3x108 km away and its mass is 3.24x105 Me.

2. With what horizontal speed must a satellite be projected at 165 km above the surface of the earth so that it will have a circular orbit about the earth? Take the earth’s radius as 6640 km. What will be the period of rotation?

3. Mars has a mean diameter of 6972 km and earth one of 13114 km. The mass of Mars is 0.11 Me. (i) How does the mean density of Mars compare with that of Earth? (ii) What is the value of g on Mars? (iii) What is the escape velocity on Mars?

4. Calculate the escape velocity of a body from the moon’s surface, if the radius of the moon is 1.7x106m and acceleration due to gravity on the moon’s surface is 1.63 m/s2.

5. A satellite wishes to orbit the earth at a height of 100 km above the surface of the earth. Determine the speed, acceleration and orbital period of the satellite. (Given Me=5.98x1024kg, Re = 6.37x106m).

6. Assuming the earth to be a sphere of uniform density 5.5 gm/cc and radius 6.4x108 cm. Calculate the value of G, the gravitational constant. Given that the acceleration due to gravity g on its surface is 988 cm/s2.

7. Calculate the escape velocity of a rocket from a surface of Mars (diameter of Mars = 6600 km, The acceleration due to gravity for the planet = 360 cm/s2).

8. The period of the moon is approximately 27.2 days (2.35x106 s). Determine the radius of the moon’s orbit and the orbital speed of the moon. (Given Me = 5.98x1024 kg, Re = 6.37x106 m). M4

9.

Four spheres from the corners of a square whose side is 2.0 cm long. What are the magnitude and direction of the net gravitational force from them, on a central sphere with mass M5 = 250 g.

M5= 250 g

M1

10.

M3

M2

The three spheres in Fig. with masses mA = 800 g, mB = 100 g and mC = 200 g have their centers on a common line, with L = 12 cm and d = 4 cm. you move sphere B L along the line until its center to center separation from C is d = 4 A B d d cm. How much work is done on sphere B (i) by you and (ii) by the net gravitational force on B due to spheres A and C.

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C