physics. gravitation - 2 session session opener how much velocity do you need to impart a stone...
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Session Opener How much velocity do you need to impart a stone such that it escapes the gravitational field of the earth?TRANSCRIPT
Physics
Gravitation - 2
Session
Session OpenerHow much velocity do you need to impart a stone such that it escapes the gravitational field of the earth?
Session Objectives
Session Objective
1. Energy Conservation in Planetary & Satellite Motion
2. Escape Velocity
3. Orbital Velocity of satellite
4. Variation of g with height & depth
5. Calculation of gravitational field due to some common systems
6. Satellite Motion and weightlessness
Energy Conservation in Planetary & Satellite Motion
Gravitational force is a conservative force. Therefore, work done by gravitational force is a independent of the path followed by a body under the force. Thus mechanical energy is conserved.
Work done by gravitational force is independent of the path followed
Escape VelocityIt is the minimum velocity with which a body must be projected vertically upward in order that it may just escape the gravitational pull of earth.
Important Fact :
Escape velocity depends on mass of planet and radiusof planet and not on mass of the object which has to begiven this velocity.
P.E.=0K.E.=0
From Conservation of EnergyK.Ei+P.Ei=K.Ef+P.E.f
e
e
e
e2
RGM2
V
0R
mGMmV
21
Re
Escape Velocity
2
ee
1K.E. mV2GM mP.E. R
ee
2GMEscape Velocity R
Orbital Velocity of satellite (Vorb)
Orbital velocity is the velocity which is given to an artificial earth’s satellite so that it may start revolving round the earth in a fixed, circular orbit.
r
mr
mVF2
a
2g rGMmF
rVa2
For motion in an orbitFa=Fg
Orbital Velocity of Satellite (Vorb)
2e2
e
GM m mVrr
GMV r
eorb
GMOrbital Velocity V r
e2ee
GM 2hg 1 for h RRR
F
MR
m
Re h
R =Re+h
Height ‘h’ above the surface of the earth.
Depth ‘d’ below the surface of the earth.
2ee
GM dg 1 RR
g With Depth and Height
2GMmF
(R h)
h 2
F GMg m (R h)
2 2 2GM g
R (1 h / R) (1 h / r)
r
t=0t=T
Hence,[Kepler’s Third Law]
Time Period of Satellite (T)
orb
Dis tance 2 r 2 rT Velocity V GMr
3rT 2 GM
22 34T rGM
Expressions of gravitational field for different bodiesGravitational field E due to a spherical shell of mass M and radius R at a point distant r from the centre.(a) When r > R
2
GMEr
(b) When r = R
2GMER
(c) When r < RE = 0
(d) When r = 0
E = 0
rR
E
Expressions of gravitational field for different bodiesGravitational field E due to a solid sphere of radius R and mass M at a point distant r from the centre.(a) When r > R (b) When r = R
2
GMEr
2
GMER
(c) When r < R (d) When r = 0
E = 0
3GMrER
E
R r
Satellite Motion and weightlessnessThe motion of satellite is based on the gravitational force. The planet appliesnecessary centripetal force by applyinggravitational force.
Inside the satellite astronauts feel weightlessness because the centrifugal force there is balanced by gravitational force.Important Facts :
•Neither the speed nor the time period depend upon the mass, size and shape of the satellite.
•Speed and time period depend upon radius (r)
Class Test
Class Exercise - 1A planet revolves in an elliptical orbit around the sun. Then out of the following physical quantities the one which remains constant is(a) velocity (b) kinetic energy(c) total energy (d) potential energy
SolutionAs there is no loss of energy in the motion of the planet in the orbits so total energy remains constant
Hence answer is (c).
Class Exercise - 2Energy supplied to change the radius of the orbit of a satellite of mass m from r to 2r is
(a) (b)
(c) zero (d) None of these
GMm4r
GMm2r
SolutionInitial energy
iGMm 1 GME m
r 2 r
GMm2r
Final energy
fGMm 1 GME m2r 2 2r
GMm
4r
Energy suppliedE = Ef – Ei
GMm4r
Hence, answer is (a).
Class Exercise - 3A planet with uniform density and radius r shrinks in size to a new radius of , but same density. What is the ratio
of the new and old escape velocities?(a) 2 : 1 (b)
(c) 1 : 2 (d)
R2
2 : 1
1: 2
Solution
eGMVR
34 RG G 43 RR R 3
So So (Ve) new : (Ve)old = 1 : 2.
eV R
Hence, answer is (c).
Class Exercise - 4The escape velocity of a body when projected vertically upward from the surface of earth is 11.2 km/s. If the body is projected in a direction making an angle of 30o with the vertical, then its escape velocity will be(a) km/s (b) km/s
(c) km/s (d) 11.2 km/s
11.23
11.22
311.2
2
Solution
Escape velocity does not depend onangle projection.
Hence, answer is (d).
Class Exercise - 5The escape speed for a projectile inthe case of earth is 11.2 km/s. Abody is projected from the surface ofthe earth with a velocity which isequal to twice the escape speed. Thevelocity of the body, when at infinitedistance from the centre of the earth, is(a) 11.2 km/s (b) 22.4 km/s
(c) (11.2) km/s (d) (11.2) km/s3 2
SolutionTotal initial energy is
2i e
GMm 1(T.E.) m(2v )R 2
GMm GMm2R R
GMm
R
Total final energy2
f1(T.E.) = 0 mv2
21mv2
2e
GMm m VR
or V = = (11.2)e2v 2
Hence, answer (d).
Class Exercise - 6R is the radius of the earth. At a height h above the surface of the earth the
acceleration due to gravity is of its
value on the surface of the earth. Then h is(a) h = 100R (b) h = 99R(c) h =10R (d) h = 9R
1100
Solutiong at surface is 2
GMgsR
At height h 2 2GM 1 GMgh
100(R h) R
or 10R = R + h h = 9R
Hence, answer is (d).
Class Exercise - 7As we go from equator to poles, the value of g(a) remains unchanged(b) decreases(c) increases(d) decreases and then increases
Solution
2 2g ́= g - R cos
cos decreases
So g´ increases.
Hence, answer is (c).
Class Exercise - 8If the earth suddenly stops rotating, then acceleration due to gravity would not change at(a) equator (b) poles(c) a latitude 45o (d) None of these
Solution
We know that2 2g ́= g - cos
At poles, 2
So g´ = g for any value of .
Hence, answer is (b).
Class Exercise - 9A body of mass m is released from the surface of earth into a tunnel passing through the centre of the earth. What is the kinetic energy of the particle as it passes through the centre?
(a) (b)
(c) (d)
3 mgr2
1 mgr2
5 mgr2
1 mgr3
SolutionPotential energy at the surface
GMmP.E.R
At the centre of earth it has 3 GMmP.E.2 R
and K.E. = 21 mv2
2GMm 3 GMm 1Then mvR 2 R 2
21 1 GMm 1mv mgR2 2 R 2
Hence, answer is (b).
Class Exercise - 10The gravitational field in a region is given
by . What is the work
done by an external agent to slowly shift a particle of mass 2 kg from point (0, 0) to point (5 m, 4 m)?(a) –180 J (b) 180 J
(c) 20 J (d) –20 J
E 10 i j N/kg
Solution
Work doneˆ ˆ ˆ ˆW mE (5i 4 j 0i 0 j)ˆ ˆ ˆ ˆ2 10(i j) (5i 4 j)
20(5 4 ) 180 Joule
Hence, answer is (b).
Thank you