Solved examples on Gravitation Example 4 In a binary star system, two stars (one of mass m and the other of 2m) distance d apart rotate about their common centre of mass. Deduce an expression for the period of revolution. Show that the ratio of their angular momenta about the centre of mass is the same as the ratio of their kinetic energies. Solution: The centre of mass C will be at a distance d/3 and 2d/3 from the masses 2m and m respectively as shown in figure 1.10. Both the stars rotate with the same angular velocity ω around C in their respective orbits. Here the gravitational force acting on each star due to other supplies the necessary centripetal force. Gravitational force on each star = G (2m) m/d 2 . Centripetal force of star (mass m) = m r ω 2 = m (2d/3)ω 2 .·. G(2m)m/d 2 = m(2d/3) ω 2 ω = √((3Gm/d 3 ) ) .·. T = 2Π/ω=2Π√((d 3 /3Gm) ) Ans. Ratio of angular momentum =(Iω) big /(Iω) small = (2m(d/3) 2 )/(m(2d/3) 2 ) = 1/2 Ans. This is same as that of their angular momenta.
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Solved examples on Gravitation
Example 4
In a binary star system, two stars (one of mass m and the other of 2m) distance d apart rotate about
their common centre of mass. Deduce an expression for the period of revolution. Show that the ratio of
their angular momenta about the centre of mass is the same as the ratio of their kinetic energies.
Solution:
The centre of mass C will be at a distance d/3 and 2d/3 from the masses 2m and m respectively as
shown in figure 1.10. Both the stars rotate with the same angular velocity ω around C in their
respective orbits. Here the gravitational force acting on each star due to other supplies the necessary
centripetal force.
Gravitational force on each star = G (2m) m/d2.
Centripetal force of star (mass m)
= m r ω2 = m (2d/3)ω2
.·. G(2m)m/d2 = m(2d/3) ω2
ω = √((3Gm/d3 ) )
.·. T = 2Π/ω=2Π√((d3/3Gm) ) Ans.
Ratio of angular momentum
=(Iω)big/(Iω)small = (2m(d/3)2)/(m(2d/3)2) = 1/2 Ans.
This is same as that of their angular momenta.
Solution:
The situation is shown in figure given above.
Following forces act on each ball
(i) Weight of the ball mg
(ii) Tension in thread T
(iii) Force of Gravitational attraction
F = G(mm/r2).
As the system is in equilibrium these three forces can be represented by the sides of a triangle
say ACP in which AC represents the weight mg, CP represents the force F and PA represents the tension
T. Hence
tan = ((r-r')/2)/l = F/mg ...... (2)
But F/mg = (Gm2/r'2)/mg ...... (1)
From equations (1) and (2)
((r-r')/2)/l = (Gm2/r'2)/mg or (r-r')mg/2 = lGm2/r'2
gr'2 (r-r') = 2l Gm. (Proved)
Example 3
A planet of mass m moves in an elliptical orbit around the Sun so that its maximum and minimum
distances from the Sun are equal to r1 and r2 respectively. Find the angular momentum of this planet
relative to the centre of the Sun.
Solution:
As the angular momentum of the planet is constant (no external torque is acting on it), we have
mv1r1 = mv2r2
or v1r1 = v2r2
Further, the total energy of the planet is also constant, hence
-GMm/r1 + 1/2 mv12 = -GMm/r2 + 1/2 mv2
2
where M is the mass of the Sun.
2
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Solved examples on Gravitation
Example 4
In a binary star system, two stars (one of mass m and the other of 2m) distance d apart rotate about
their common centre of mass. Deduce an expression for the period of revolution. Show that the ratio of
their angular momenta about the centre of mass is the same as the ratio of their kinetic energies.
Solution:
The centre of mass C will be at a distance d/3 and 2d/3 from the masses 2m and m respectively as
shown in figure 1.10. Both the stars rotate with the same angular velocity ω around C in their
respective orbits. Here the gravitational force acting on each star due to other supplies the necessary
centripetal force.
Gravitational force on each star = G (2m) m/d2.
Centripetal force of star (mass m)
= m r ω2 = m (2d/3)ω2
.·. G(2m)m/d2 = m(2d/3) ω2
ω = √((3Gm/d3 ) )
.·. T = 2Π/ω=2Π√((d3/3Gm) ) Ans.
Ratio of angular momentum
=(Iω)big/(Iω)small = (2m(d/3)2)/(m(2d/3)2) = 1/2 Ans.
This is same as that of their angular momenta.
4
Example 5
Imagine a planet whose diameter and mass are both one half of those of Earth. The day's temperature
of this planet's surface reaches upto 800K. Find whether oxygen molecules are possible in the
Which, on substituting the given value, is 3/5 GM2/R = 3/5 g MR
= 1.5 × 1032 J.
Example 9
A satellite revolving close to the surface of Earth from west to east, appears over a certain point at the
equator every 11.6 hours. If radius of Earth is 6400 km. calculate the mass of Earth.
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Solution:
Since the rotation of Earth is also west to east the apparent angular velocity of the satellite is equal to
ωSE = ωS - ωe,
Where ωS is the angular velocity of the satellite w.r.t. an irrotational frame of reference fixed on the
axis of Earth's rotation and ωe is the angular velocity of Earth w.r.t. the same frame of reference.
.·. ωS = ωSE + ωe
now (GMem)/R2 m Rω2,
where R = radius of Earth as the satellite is rotating close to its surface and M e is the mass of Earth and
m is the mass of satellite.
.·. M = (R3 ω2)/G = R3/G [2Π/TSE + 2Π/TE ]2.
where TSE is the apparent time period of satellites revolution and TE = 24 hrs. is the time period of
Earth's rotation.
Substituting the values
M = 6.0 × 1024 kg.
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Two questions on gravitation were included in the IIT-JEE 2010 question paper. They are given below with solution. The first question is a ‘single correct choice type’ (single answer type multiple choice) question where as the second question is an ‘integer type’ question which has a single digit integer ranging from 0 to 9 as answer.
(1) A thin uniform annular disc (see figure) of mass M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P on its axis to infinity is(A) (2GM /7R) (4√2 – 5)(B) – (2GM /7R) (4√2 – 5)(C) GM /4R(D) (2GM /5R) (√2 – 1)The work required to take a unit mass from infinity to point P is the gravitational potential at P and it will be negative. Therefore, the work required to take a unit mass from point P to infinity will be numerically equal to the gravitational potential at P but it will be positive.The annular disc can be considered to be made of a large number of concentric rings of radii ranging from 3R to 4R. Consider one such ring of radius r and small thickness dr. Its mass dm is given bydm = [M/π(16R2 – 9R2)] 2πr dr = 2Mrdr/7R2
If the rim of the ring is at distance x from the point P, the gravitational potential (dV) at P due to the ring is given by dV = – Gdm/x = – (2GM/7R2)(rdr/x) But x = (16R2 + r2)1/2 and so we have dx= (½ )×(16R2 + r2)–1/2 ×2rdrOr, dx = rdr/(16R2 + r2)1/2 = rdr/xTherefore dV = – (2GM/7R2)dxThe gravitational potential at P due to the entire annular disc is given byV = ∫ dV = – ∫(2GM/7R2)dx = – (2GM/7R2)[x] The integration is between the appropriate limits of x, which are (16R2 + 9R2)1/2 and (16R2 + 16R2)1/2, on substituting the values r = 3R and r = 4R respectively in the expression, x = (16R2 + r2)1/2. Therefore we haveV = – (2GM/7R2)[(16R2 + 16R2)1/2 – (16R2 + 9R2)1/2] = – (2GM/7R2)[R√(32) – R√(25)] Thus V = – (2GM/7R) (4√2 – 5)The negative sign is to be dispensed with to obtain the work required to take a unit mass from point P to infinity and the correct choice is (2GM/7R) (4√2 – 5). (2) Gravitational acceleration on the surface of a planet is (√6/11)g, where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 kms–1, the escape speed on the surface of the planet in kms–1
will be :[This is is an ‘integer type’ question. You have to work out this question and the single digit integer answer is to be shown in the Objective Response Sheet (ORS) by darkening the appropriate bubble].
We have vesc = √(2gR) where g is the acceleration due to gravity and R is the radius of the planet (or earth or star). Therefore vesc α √(gR).If gp and ge are the accelerations due to gravity on the surface of the planet and the earth, Rp and Re are their radii, and gp and ge the escape velocities respectively, we havevp/ve = √( gpRp)/ √( geRe) = √[( gp/ge)(Rp/Re)] ………(i)Since the gravitational acceleration on the surface is given by g = GM/R2 where G is the gravitational constant and M is the mass of the planet (or earth or star), we can writeg = G[(4/3)πR3ρ] /R2 where ρ is the average mass density of the planet (or earth or star).Therefore g α Rρ and gp/ge = (Rp/Re)(ρp/ρe).From this Rp/Re = (gp/ge)/(ρp/ρe).But gp/ge = √6/11 and ρp/ρe = 2/3 as given in the question. Therefore we haveRp/Re = (√6/11)×3/2Substituting these in Eq(i),vp/ve = √[(√6/11) (√6/11)×3/2] = (√6/11) ×√(3/2) = 3/11Since ve = 11 kms–1, the escape speed on the surface of the planet in kms–1 will bevp = 11×(3/11) = 3 kms–1 The answer integer is 3.
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Solved examples of Gravitation
Example 1
The time period of Moon around the Earth is n times that of Earth around the Sun. If the ratio of
the distance of the Earth from the Sun to that of the distance of Moon from the Earth is 392, find the
ratio of mass of the Sun to the mass of the Earth. (Assume that the bodies revolve in circular orbits)
Solution:
The time period Te of Earth around Sun of mass Ms is given by
Te2 = 4Π2/GMe × re
3, ...... (1)
where re is the radius of the orbit of Earth around the Sun.
Similarly, time period Tm of Moon around Earth is given by
Tm2 = 4Π2/GMe × re
3, ...... (2)
where rm is the radius of the orbit of Moon around the Sun.
Dividing equation (1) by (2), we get
(Te/Tm)2 = (Me/Ms) (re/rm)3
.·. (Ms/Me) = (Tm/Te )2× (re/rm)3 ...... (3)
Substituting the given values, we get
(Ms/Me ) = (392)3 n2 Ans.
Example 2
Two balls of mass m each are hung side by side, by two long threads of equal length l. If the distance
between upper ends is r, show that the distance r' between the centres of the ball is given by gr'2 (r-r')
= 2l G M.
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Solution:
The situation is shown in figure given above.
Following forces act on each ball
(i) Weight of the ball mg
(ii) Tension in thread T
(iii) Force of Gravitational attraction
F = G(mm/r2).
As the system is in equilibrium these three forces can be represented by the sides of a triangle
say ACP in which AC represents the weight mg, CP represents the force F and PA represents the tension
T. Hence
tan = ((r-r')/2)/l = F/mg ...... (2)
But F/mg = (Gm2/r'2)/mg ...... (1)
From equations (1) and (2)
((r-r')/2)/l = (Gm2/r'2)/mg or (r-r')mg/2 = lGm2/r'2
gr'2 (r-r') = 2l Gm. (Proved)
Example 3
A planet of mass m moves in an elliptical orbit around the Sun so that its maximum and minimum
distances from the Sun are equal to r1 and r2 respectively. Find the angular momentum of this planet
relative to the centre of the Sun.
Solution:
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As the angular momentum of the planet is constant (no external torque is acting on it), we have
mv1r1 = mv2r2
or v1r1 = v2r2
Further, the total energy of the planet is also constant, hence
-GMm/r1 + 1/2 mv12 = -GMm/r2 + 1/2 mv2
2
where M is the mass of the Sun.
Example 4
In a binary star system, two stars (one of mass m and the other of 2m) distance d apart rotate about
their common centre of mass. Deduce an expression for the period of revolution. Show that the ratio of
their angular momenta about the centre of mass is the same as the ratio of their kinetic energies.
Solution:
The centre of mass C will be at a distance d/3 and 2d/3 from the masses 2m and m respectively as
shown in figure 1.10. Both the stars rotate with the same angular velocity ω around C in their
respective orbits. Here the gravitational force acting on each star due to other supplies the necessary
centripetal force.
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Gravitational force on each star = G (2m) m/d2.
Centripetal force of star (mass m)
= m r ω2 = m (2d/3)ω2
.·. G(2m)m/d2 = m(2d/3) ω2
ω = √((3Gm/d3 ) )
.·. T = 2Π/ω=2Π√((d3/3Gm) ) Ans.
Ratio of angular momentum
=(Iω)big/(Iω)small = (2m(d/3)2)/(m(2d/3)2) = 1/2 Ans.
This is same as that of their angular momenta.
Example 5
Imagine a planet whose diameter and mass are both one half of those of Earth. The day's temperature
of this planet's surface reaches upto 800K. Find whether oxygen molecules are possible in the
Which, on substituting the given value, is 3/5 GM2/R = 3/5 g MR
= 1.5 × 1032 J.
Example 9
A satellite revolving close to the surface of Earth from west to east, appears over a certain point at the
equator every 11.6 hours. If radius of Earth is 6400 km. calculate the mass of Earth
Solution:
Since the rotation of Earth is also west to east the apparent angular velocity of the satellite is equal to
ωSE = ωS - ωe,
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Where ωS is the angular velocity of the satellite w.r.t. an irrotational frame of reference fixed on the
axis of Earth's rotation and ωe is the angular velocity of Earth w.r.t. the same frame of reference.
.·. ωS = ωSE + ωe
now (GMem)/R2 m Rω2,
where R = radius of Earth as the satellite is rotating close to its surface and M e is the mass of Earth and
m is the mass of satellite.
.·. M = (R3 ω2)/G = R3/G [2Π/TSE + 2Π/TE ]2.
where TSE is the apparent time period of satellites revolution and TE = 24 hrs. is the time period of
Earth's rotation.
Substituting the values
M = 6.0 × 1024 kg.
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1. A solid sphere of uniform density and radius 4 units is located with its centre at origin O of coordinates. Two units of equal radii 1 unit, with their centre at A
(-2,0,0) and B (2,0,0) respectively are taken out of the solid leaving behind spherical cavities as shown in Fig. Choose the incorrect statement from the following.
a)The gravitational force due to this object at origin is zero.
b)The gravitational force at point B (2,0,0)
c)The gravitational potential is the same at all points of the circle Y2 + Z2 = 36
d)The gravitational potential is the same at all points of the circle Y2 + Z2 = 4
Ans. (b)
The distance of each cavity from the centre O is the same. The two cavities are symmetrical with respect to the centre O and the mass of the sphere being concentrated at the centre O, the gravitational force due to the sphere is zero at the centre. Hence choice (a) is correct. Similarly, the gravitational potential is the same at all points of the circle Y2+Z2 = 36 whose radius is 6 units and at all points of the circle Y2+Z2 = 4. So choices (c) and (d) are also correct.
2. If the distance between the earth and the sun were half its present value, the number of days in a year would have been
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a)64.5 (b) 129
b)182.5 (d) 730
Ans. (b)
According to Kepler’s law of periods,
T1/T2 = (R1/R2)3/2 = [ R1/ (R1/2)]3/2
= (2)3/2 = 2√2
T2 = T1/2√2 = 365 days/2√2 = 129 days
3. An artificial satellite moving in a circular orbit around earth has a total (kinetic+potential) energy E0. Its potential energy is
4. A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth. Which of the following statement is correct ?
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(a) The acceleration of S is always directed towards the centre of the earth.
(b) The angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant.
(c) Total mechanical energy of S remains constant.
(d) The linear momentum of S remains constant in magnitude.
Ans. (a)
For elliptical orbit, the earth is at one focus of the ellipse. For spherical bodies, the gravitational force is central. Hence statement (a) is correct.
5. Two objects of masses m and 4m are at rest at infinite separation. They move towards each other under mutual gravitational attraction. Then at a separation r, which of the following is true?
(a) The total energy of the system is not zero.
(b) The force between them is not zero.
(c) The centre of mass of the system is at rest.
(d) All the above are true
Ans. (d)
At a finite separation the total kinetic energy of the system of masses and the force between them are both finite. Since the two masses are at rest initially and there is no external force, the centre of mass cannot move. Hence the correct choice is (d)
6. A satellite is launched in to a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius 1.01R. The period of the second satellite is longer than that of the first by approximately
0.5% (b) 1.0%
1.5% (d) 3.0%
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Ans. (c)
We know according to Kepler’s law of periods, T2 = KR3 where k is a constant. Taking logarithm of both sides, we have
2 log T = log k + 3 log R
Differenting, we get
2 (dT / T) = 0 + (3 dR / R)
= 3/2 x [(1.01R – R) / R] x 100 = 1.5%
so correct choice is (c)
7. A simple pendulum has a time period T1 when on the earth surface, and T2 when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of T2/T1 is
(a) 1 (b) √2
(c) 4 (d) 2
Ans. (d)
The acceleration due to gravity at a height h above the surface of the earth is given by
g2 = g1 [R / R+h]2
where g1 is the value at the surface of the earth
T2 = 2π √l/g2 and T1 = 2π √l/g1
T2 / T1 = √g1/g2 = R+h / R = R+R / R = 2 (h=R)
8. An ideal spring with spring constant is k is hung from the ceiling and a block and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is
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(a) 4 Mg/k (b) 2 Mg/k
(c) Mg/k (d) Mg/2k
Ans. (b)
Let x be the extension in the spring when it is loaded with mass M. The change in gravitational potential energy = Mgx. This must be the energy stored in the spring which is given by ½ kx2. Thus ½ kx2 = Mgx or x = 2mg / k
Hence the correct choice is (b)
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Gravitation Practice Problems - Solutions
1. All planets orbit the sun in ellipses, not perfect circles which means that there is a point in each planet’s orbit when it is closest to the sun and when it is farthest from the sun. Comets orbit the sun in much the same way, but the distance differences are much more drastic. As Halley’s Comet orbits the sun, its closest distance is 5*1010m and its farthest distance is 4.5*1012m.
a. Without doing any calculations, compare the gravitational forces between the comet and the sun at both the closest and farthest points in the orbit?
The farthest distance is 90 (4.5*1012/5*1010 = 90) times greater than the closest distance. Therefore, the force becomes (1/90)2 [or 0.000123]
times the original size.
b. If we assume the comet has a mass of 3*1014kg, calculate the two gravitational forces.FG closest = G*3*1014*2*1030/ (5*1010)2 = 1.6*1013 N
FG farthest = G*3*1014*2*1030/ (4.5*1012)2 = 1.98*109 N
2. How do we know that Jupiter has a force exerted on it?It moves in a circular orbit
3. What exerts the force on Jupiter?The Sun exerts a gravitational force on Jupiter
4. What do we mean when we say that Jupiter falls?Jupiter falls from its straight line or tangential path onto the curved path of its orbit
5. How do gravitational force and distance compare?Gravitational force is proportional to the inverse square of the distance
6. At what height above the earth would a 300kg satellite have to orbit in order to experience a gravitational force half as strong as it is on the Earth’s surface?
FG on the Earth’s surface = G*300*6*1024/(6.4*106)2 = 2931 N½ * 2931 = 1465.5 = G*300*6*1024/d2
d2 = G*300*6*1024/1465.5 = 8.19*1013 d = 9.05*106m = x +REARTH
x = 9.05*106 – 6.4*106 = 2.65*106m above the surface of the Earth
7. The acceleration due to gravity on Venus is 0.89 that of Earth. If the radius of Venus is 6.05*106m, what is Venus’ mass?
FG = 0.89*9.8 = 8.72= G*1*MVENUS/(6.05*106)2
MVENUS = 8.72*(6.05*106)2/(G*1) = 4.79*1024 kg
8. Which exerts a greater force, the Sun on Venus or Venus on the Sun?They both exert the same force on one another.
By Newton’s Third Law, we know this must be true
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Newton’s Law of Gravitation
Newton's Law of Gravitation states that "Every particle in the universe attracts every other particle with
a force that is directly proportional to the product of their masses and inversely proportional to the
square of the distance between them".
Consider two bodies A and B of masses mA and mB, attracting each other with forces AB (force on A due
to B) and BA (force on B due to A), respectively.
Then, AB = - BA
and
AB = | AB| = magnitude of the attractive force
= G(mA mB)/d2
where d is the distance between them. G is a universal constant known as Universal Gravitational
constant. Its value was first measured by Cavendish and is now known to be:
G = 6.62726 × 10-11 N-m2/kg2
"The space around a body within which its force of gravitational attraction is perceptible (by any other
body in this space) is called its gravitational field."
The intensity E, , of the gravitational field of a mass 'm' at a point at distance 'r' from it is the force
experienced by a unit mass placed at this point in the field. (Assuming that the presence of unit mass
does not affect the gravitational field of the mass m)
Thus, E = (-m/r2)G (the negative sign is because the intensity of the field is directed towards the
mass and away from the unit mass).
If 1, 2, 3 ...... are the gravitational forces acting on the particle A due to particles P1, P2, P3, ......,
respectively, then net gravitational force on a particle A due to particles P1, P2, ......, Pn is given by:
= 1 + 2 +........+ n
Illustration:
Two Lead balls of radius 10 cm and 1 cm are placed with their centres 1 metre apart. Calculate the
force of attraction between them. The density of Lead is 5.51 × 103 kg/m3.
Newton’s Law of Gravitation
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Solution:
Density of balls, ρ = 5.51× 103 kg/m3
Force of attraction between them, F = Gm1m2/r2
where m1 = 4/3Πr23 × ρ = 5.51 × 103 × 4/3 Π10-3 ≈ 23 kg
and m2 = 4/3Πr23 × ρ ≈ 23 × 10-3 kg
G = 6.67 × 10-11 Nm2 kg-2
.·. F = 6.67 × 10-11 × 21 × 21 × 10-3N = 3.5 × 10-11 N Ans.
Important points
← The gravitational force is an attractive force.
← The gravitational force between two particles does not depend on the medium.
← The gravitational force between two particles is along the straight line joining the particles
(called line of centers).
Illustration:
Three identical particles, each of mass m, are placed at the three corners of an equilateral triangle of
side 'a'. Find the force exerted by this system on another particle of mass m placed at
(a) the mid-point of a side
(b) the centre of the triangle
Solution:
(a) Using the principle of superposition
= A + B + C
When the particle is placed at the mid point of a side (at P), C = - B, and they cancel each other.
Hence, force experienced by the particle, = A
| | = | FA | = Gmm/(AP)2 = Gm2/(a sin 600)2
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(b) If the particle is placed at the centre of the triangle, the net force on the particle P due to particles
placed at the corners A, B and C will be zero.
Hence, = A + B + C= 0.
Gravitational Field and Intensity
The space around a body where the gravitational force exerted by it can be experienced by any other
particle is known as the gravitational field of the body. The strength of this gravitational field is referred
to as intensity, and it varies from point to point.
Consider the gravitational field of a particle of mass m located at the origin (O).
Suppose that a test particle of mass m0 is placed at the point P(x, y, z). The force of gravitational
attraction exerted on the test particle is given by,
g = (GMmo/r2)
where the position vector = r,
r = OP = | | = | | = r
and the unit vector, = / r
The intensity of this gravitational field at a point (P) is given by the force per unit mass on a test
particle kept at P, i.e.
= g/mo
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where is the gravitational intensity and g is the gravitational force acting on the mass m0. The
gravitational field is, therefore, a vector field.
The gravitational field at P due to a particle of mass m kept at the point O (origin) is given by
= g/mo = {-(Gmmo/r2) } * 1/mo = Gm/r2
where = xi + yj + zk represents the position vector of the point P with respect to the source at the
origin and = / r represents the unit vector along the radial direction.
The superposition principle extends to gravitational field (intensities) as well:
= 1 + 2 + 3 +....+ n
where 1, 2,.... n are the gravitational field intensities at a point due to particle 1, 2, ......, n respectively.
For a continuously distributed mass, the formula changes to =∫d , where d gravitational field
intensity due to an elementary mass dm.
Gravitational Field and Intensity
The gravitational field of a ring on its axis
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Let us consider a ring of mass M in the plane perpendicular to the plane of the paper. We want to find
the gravitational field on its axis at a distance x.
Consider a differential length of the ring of mass dm.
dE = Fdm/z2
The Y-components of the fields due to diametrically opposite elements cancel each other. Thus, the X-