Class 11 - Physics - Gravitation
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?
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 (r2 – r1) for the
difference of potential energy between two points r2 and r1 distance away from the centre of the earth.
g = (1-(2h)/ (Re)) g Where,
Re= Radius of the Earth g = Acceleration due to gravity on the surface of the Earth.
It is clear from the given relation that acceleration due to gravity decreases with an increase in height.
gd = (1-(d/Re)) g
It is clear from the given relation that acceleration due to gravity decreases with an increase in depth.
Acceleration due to gravity of body of mass m is given by the relation:
g = (GM)/ (R2)
G = Universal gravitational constant
M = Mass of the Earth
R = Radius of the Earth
It is clear that acceleration due to gravity is independent of the mass of the body.
Gravitational potential energy of two points r2 and r1 distance away from the centre of the Earth is respectively given by:
V (r1) = - (GMm)/ (r1)
V (r2) = - (GMm)/ (r2)
Therefore difference in potential energy, V = V (r2) - V (r1)
=-GmM ((1/ r2) – (1/ r1))
Hence, this formula is more accurate than the formula mg (r2– r1).
Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?
Radius of orbit of earth = re = 1 AU
Radius of orbit of planet = rp
Also Time taken by the earth to complete one revolution around the sun,
Te = 1 year
Time taken by the planet to complete one revolution around the sun,
Tp = (1/2) Te
= (1/2) year
From Kepler’s third law of planetary motion,
(rp/ re)3 =( Tp/ Te)2
(rp/ re) = (Tp/ Te) (2/3)
= ((1/2)/ (1)) (2/3)
= (0.5) (2/3)
Therefore, the orbital radius of the planet will be 0.63 times smaller than that of the Earth.
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.
Orbital period of Io, TIo = 1.769 days = 1.769 x 24 x 60 x 60 sec
Orbital radius of Io, Rlo = 4.22 x 108 m
Satellite Io is revolving around the Jupiter
Mass of the satellite is given as:
MJ = (4 π2 RIo3)/ (GTIo2)
MJ = Mass of Jupiter
G = Universal gravitational constant
Orbital period of the Earth, Te = 365.25 days = 365.25 x 24 x 60 x60s
Orbital radius of the Earth,
Re = 1 AU = 1.496 x 1011 m
Mass of the Sun Ms = (4π 2 Re3)/ (GTe2) … (ii)
Therefore, (Ms/ MJ) = (4 π2 Re3)/ (GTe2) x (GTIo2)/ (4π 2 RIo3)
= ((1.769 x 24 x 60 x 60)/ (365.25 x 24 x 60 x60)) 2 x ((1.496 x 1011)/ (4.22 x 108))3
Therefore, (Ms/ MJ) ≈ 1000
Ms ≈ 1000 MJ
This shows the mass of Jupiter is (1/1000th) the mass of the sun.
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.
Mass of the galaxy, M = 2.5 × 1011 solar mass
=2.5 × 1011 x (2 x 1030)
=5 x 1041kg
Radius of the orbit of star, r= 50000ly
=50000 x 9.46 x 1015m
=4.73 x 1020m
Therefore time period of the star T =2 π √ (r3)/ (GM)
=2 π √ (4.73 x 1020)3/ (6.67 x 10-11 x 5 x 1041)
=1.12 x 1016 sec
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 required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy
required to project a stationary object at the same height (as the satellite) out of earth’s influence.
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?
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.
Which of the following symptoms is likely to afflict an astronaut in space?
In the following two exercises, 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) 0.
Correct option: (iii).
Gravitational potential (V) is constant at all points inside a spherical shell.
Therefore gravitational potential gradient (-dV/dr) at all points inside a spherical shell is zero.
Since gravitational intensity E (-dV/dr), it shows gravitational intensity is 0 at all points inside the spherical shell.
When we remove the upper part of the spherical shell, we will get hemispherical shell.
As the gravitational intensity at centre C is 0, the direction of gravitational intensity must be downwards. Therefore option (iii) c is correct.
For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.
Correct option: - (ii) e
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.
A rocket is fired 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).
Mass of Sun M = 2× 1030 kg
Mass of the earth m =6× 1024 kg
Distance between Sun and Earth r = 1.5 × 1011 m
Consider a point R, where the gravitational force on the rocket due to earth = gravitational force on the rocket due to sun.
Distance of point R from the Earth = x
Therefore, (Gm)/ (r2) = (GM)/(r-x) 2
(r-x) 2/(x2) = (M/m)
= (2 x 1030)/ (6 x 1024)
= (106)/ (3)
=(r-x)/(x) = ((10)3/ (√3)
(r/x) = ((10)3/ (√3) + 1
≈ (10)3 / (√3)
x = (√3) r/ (10)3
= (1.732 x 1.5 x 1011)/ (103) m
=2.6 x 108 m
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.
Radius of the earth orbit R= 1.5 × 1011
Time period of the earth around the sun = 365 years
= 365 x 24 x 60 x 60 s
Gravitational constant G = 6.67 x 10-11 Nm2kg-2
Time period of the satellite = T = 2π√(R+h) 2/ (GM)
Or Mass of the sun, M = 4 π2(R+h) 3 / (GT2)
= (4 π2 x (1.5 × 1011)3)/ (6.67 x 10-11) x (365 x 24 x 60 x60)2
=2.01 x 1030 kg
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?
Distance of the Earth from the Sun, re = 1.5 × 108 km = 1.5 × 1011 m
Time period of the Earth = Te
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 = (4π r3/GM) 1/3
For Saturn and Sun, it can be written as:
(rs) 3/ (re)3 = (Ts) 2/ (Te) 2
rs = re (Ts/ Te)(2/3)
=1.5 × 1011 (29. 5 Te/Te) (2/3)
=1.5 × 1011 x (29. 5) (2/3)
= 1.5 × 1011 x 9.55
=14.32 x 1011m
Hence, the distance between Saturn and the Sun = 14.32 x 1011m
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?
Weight of the body = 63N
Height h = (Re/2) where Radius of the earth = Re
Acceleration due to gravity at height h from the Earth’s surface is as:
g = Acceleration due to gravity on the Earth’s surface
g’ = (g)/(1+ (Re/2xRe))2
= (g)/ (1+ (1/2)) 2
= (4/9) g
Weight of a body of mass m at height h is given as:
W’ = mg’
=mx (4/9) g
= (4/9) mg
= (4/9) x 63
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?
Weight of a body of mass m at the Earth’s surface, W = mg = 250 N
Where g = acceleration due to gravity on earth’s surface, mass of the body = m and weight of the body = W
Depth, d = (1/2) Re
Where, Re = Radius of the Earth
Acceleration due to gravity at depth =g’ and is given as:
g' = (1- (d/ Re))g
= (1- (Re)/ (2x Re)) g
= (1/2) g
Weight of the body at depth d,
W’ = mg’
= m x (1/2) g
= (1/2) W
= (1/2) x 250
A rocket is fired 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 kg–2.
Velocity of the rocket, v = 5 km/s = 5 × 103 m/s
Mass of the Earth, Me = 6.0 x 1024 kg
Radius of the Earth, Re = 6.4 x 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)
By applying law of conservation of energy,
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 + h - Re)/ (Re (Re + h)))
(1/2) v2 = ((GMe h)/ (Re (Re + h))) x (Re/ Re)
(1/2) v2 = (g Re h)/ (Re + h)
Where g (acceleration due to gravity on earth’s surface)
= (GMe /Re2)=9.8m/s2
Therefore, (v2 Re) = h (2gReh)
h= (Re v2)/ (2gRe - v2)
= (6.4 x 106 x (5 × 103)2)/ (2 x 9.8 x 6.4 x 106 – (5x 103)2)
h = (6.4 x 25 x 1012)/ (100.44 x 106)
h = 1.6 x 106 m
Height achieved by the rocket with respect to the centre of the Earth
= (Re + h)
= (6.4 x 106 + 1.6 x 106)
=8.0 x 106 m
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.
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) m v2esc
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) m v2esc = (1/2) mvf2
vf = √ (vp2 - v2esc)
=√ (3vesc) 2 – (v2esc)
=√ (8) vesc
=√ (8) x 11.2
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.
Mass of the Earth, Me = 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
Potential energy of the satellite when revolving = (-GMem)/ (Re + h)
Kinetic energy of the satellite when revolving = (1/2) mv2
Orbital velocity of the satellite, v=√ (GMe)/ (Re +h)
Total energy of the orbiting satellite = (1/2) mv2 + (-GMem)/ (Re + h)
=- (1/2) (GMem)/ (Re +h)
(-) ive sign indicates that the satellite is bound to the earth. This energy is known as Bound energy.
Energy to be imparted to put the satellite out of earth’s gravitational influence
= – (Bound energy)
= (1/2) (GMem)/ (Re +h)
= (1/2) x (6.67 × 10–11 x 6.0 × 1024 x 200)/ (6.4 × 106 + 0.4 ×106)
= (1/2) x (6.67 x6 x 2 x 10)/ (6.8 x 106)
= (5.9 x 109) J
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).
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 v = 0 total energy of two stars separated at distance r
= (-GMm)/(r) + (1/2) mv2
= (-GMm)/(r) + 0 … (i)
When the stars are about to collide:
Velocity of the stars = v
Distance between the centres of the stars will be twice the radius of a star i.e.
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,
Mv2 - (GMm)/ (2R) = - (GMM)/(r)
v2 = - (GM)/(r) + (GM)/ (2R)
=GM ((-1/r) + (1/2R))
=6.67 x 10-11 x 2 x 1030[(-1/1012) + (1/2) x107)]
=13.34 x 1019 [-10-12 + (5 x 10-8)]
=-6.67 x 1012
v= √ (6.67 x 1012)
v= (2.58 x 106) m/s
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 centres of the spheres?
Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?
Gravitational Constant G = 6.67 x 10-11 Nm2kg-2
Mass of spheres = 100kg
Radius of each sphere = 0.10m
Distance between two spheres d = 1.0m
Let mid-point between two spheres = T
Therefore TA = TB =r = (d/2) = 0.5m
Force due to gravitational field at T due to sphere A = (GM)/ (0.5)2
Force due to gravitational field at T due to sphere B = (GM)/ (0.5)2
At mid- point gravitational force will be zero because gravitational force exerted by each sphere will act in opposite directions.
Gravitational potential at mid-point due to two spheres will be:
= (-GM/(r) + (GM)/(r) = - (2GM/r)
= - (6.67 x 10-11 x 100 x2)/ (0.5)
=-2.668 x 10-8 Jkg-1
Any object placed at mid-point will be in equilibrium state, but the equilibrium is unstable.
If the objects are displaced from mid-point towards either sphere, the object will not return to its initial position of equilibrium.