What is the change in mass of a body when taken 64 km below the surface of the earth take radius of the earth as 6400 km?

CBSE 9 - Physics

Asked by kalpanatiwaritiwari7030 | 22 Dec, 2021, 07:35: PM

What is the change in mass of a body when taken 64 km below the surface of the earth take radius of the earth as 6400 km?
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1. Newton's Law of Gravitation:

(i) Force of attraction between two point masses is F=Gm1m2r2, where G=6.67×10-11 N m2 kg-2 is the universal gravitational constant and r is the separation between the masses m1 and m2.

(ii) It is directed along the line joining point masses.

(iii) It is a conservative force field ⇒ mechanical energy is conserved.

(iv) It is a central force field ⇒ angular momentum is conserved.

2. Newton's Law of Gravitation in Vector Form:

What is the change in mass of a body when taken 64 km below the surface of the earth take radius of the earth as 6400 km?

(i) F→12=Gm1m2r3-r→12

(ii) F→21=Gm1 m2r3r→12

(iii) F→12=-F→21

3. Gravitational Field E→:

It is defined as gravitational force per unit mass.

(i) It is a vector quantity.

(ii) E→=F→m

Due to a point mass, E→=F→m=-GMr3r→

(iii) Unit: N kg-1 or m s-2

(iv) Dimensions: M0L1T-2

4. Gravitational Potential V:

At a point in a gravitational field, potential V is defined as the negative of work done by the gravitational field W per unit mass m in shifting a test mass from some reference point (usually at infinity) to the given point.

(i) It is a scalar quantity.

(ii) Unit: J kg-1 or m2 s-2

(iii) Dimensions: M0L2T-2

(iv) V=-W ∞rm=-∫∞rF→.dr→m=-∫∞rE→.dr→

(v) E=-dVdr

(vi) Negative gradient of potential gives intensity of field or potential is a scalar function of position whose space derivative gives intensity. The negative sign indicates that the direction of intensity is in the direction where the potential decreases.

(vii) ∆V=-∫E→·dr→=-∫-GMr2dr

(viii) Gravitational potential due to a point mass at a distance r from the mass, V=-GMr

5. Gravitational Field and Gravitational Potential of Continuous Bodies:

(i) Point mass, V=-GMr, E=-dVdr=-GMr2

(ii) Uniform Ring of mass M, radius a;

V=-GMx or -GM a2+r21/2 and E=-GMra2+r23/2r^ or E=-GMcosθx2

The gravitational field is maximum at a distance, r=±a/2 and Emax=-22GM33a2

What is the change in mass of a body when taken 64 km below the surface of the earth take radius of the earth as 6400 km?

(iii) Uniform Thin circular disc (Mass =M, Radius =a), 

V=-2GMa2a2+r212-r and E=-2GMa21-rr2+a212=-2GMa21-cosθ

(iv) Solid sphere:

(a) If point P is inside the sphere r≤a,

V=-GM2a33a2-r2 and E=-GMra3, and at the centre r=0, V=-3GM2a and E=0

(b) If point P is outside the sphere r≥a,

V=-GMr and E=-GMr2

What is the change in mass of a body when taken 64 km below the surface of the earth take radius of the earth as 6400 km?

(v) Uniform thin spherical shell:

(a) If point P is inside the shell r≤a,

V=-GMa (constant) and E=0

(b) If point P is outside shell r≥a,

V=-GMr and E=-GMr2

What is the change in mass of a body when taken 64 km below the surface of the earth take radius of the earth as 6400 km?

6. Gravitational Potential Energy of a System of Particles:

It is the work done by an external force against the internal gravitational field to form a system of particles in a particular configuration from infinite separation without any change in their kinetic energies.

(i) Gravitational potential energy of two-particle system: U12=-Gm1m2r where r is the separation between the masses m1 and m2.

(ii) Gravitational potential energy of a system of particles: U=U12+U13+.......+U1n+U23+U24+.......+U2n+.......

7. Gravitational Self Energy: The work done by an external agent in assembling a system of particles from infinitesimal particles which are initially infinite distance apart.

(i) Gravitational self-energy of a uniform solid sphere: Usphere=-35GM2R

(ii) Gravitational self-energy of a uniform hollow sphere (spherical shell): Ushell=-12GM2R

8. Mass and Density of Earth:

Newton’s law of gravitation can be used to estimate the mass and density of the earth.

(i) M=gR2G=9.8×6.4×10626.67×10-11=5.98×1024kg ≈1025 kg

(ii) ρ=3g4πGR=3×9.84×3.14×6.67×10-11×6.4×106=54784 kg m-3

9. Acceleration due to Gravity:

The force of attraction exerted by the earth on a body is called the gravitational pull of gravity.

(i) Variation in g due to the shape of the earth: The equatorial radius is about 21 km longer than the polar radius, from g=GMR2,

(a) gegp=Rp2Re2

(b) Since, Requator>Rpole,  ∴ gpole>gequator

(c) gp=ge+0.018 m s-2

(ii) Variation in g with height:

What is the change in mass of a body when taken 64 km below the surface of the earth take radius of the earth as 6400 km?

(a) g'=gRR+h2=gR2r2 ,

(b) If h≪R, g'=gRR+h2=g1+hR-2=g1-2hR

(c) If h≪R then decrease in the value of g with height:

Absolute decrease Δg=g-g'=2hgR

Fractional decrease Δgg=g-g'g=2hR

Percentage decrease Δgg×100%=2hR×100%

(iii) Variation in g With Depth:

g'=g1-dR

What is the change in mass of a body when taken 64 km below the surface of the earth take radius of the earth as 6400 km?

(iv) Variation in g due to rotation of the earth:

What is the change in mass of a body when taken 64 km below the surface of the earth take radius of the earth as 6400 km?

(a) If the body of mass m lying at point P, whose latitude is θ, then due to rotation of earth its apparent weight can be given by mg'→=mg→+F→c, here Fc=mω2r=mω2Rcosθ

(b) g'=g-ω2Rcos2θ

10. Inertial and Gravitational Masses:

(i) Inertial mass: It is the mass of the material of the body, which measures its inertia. If an external force F acts on a body of mass mi, then according to Newton’s second law of motion, F=mia⇒ mi=Fa

(ii) Gravitational mass: It is the mass of the material of the body, which determines the gravitational pull acting upon it. If M is the mass of the earth and R is the radius, then gravitational pull on a body of mass mgis given by F=GMmgR2or mg=FGMR2=FE

Here, mg is the gravitational mass of the body, if E=1 then mg=F. Thus the gravitational mass of a body is defined as the gravitational pull experienced by the body in a gravitational field of unit intensity,

11. Satellite Velocity (or Orbital Velocity):

(i) v0=GMeRe+h12=gRe2Re+h12

(ii) If h≪Re then v0=gRe

∴ v0=9.8×6.4×106=7.92×103 m s-1=7.92 km s-1

(iii) Time period of Satellite

T=2πRe+hgRe2Re+h12=2πReRe+h3 g12

12. Energy of a Satellite:

(i) Potential energy, U=-GMemr

(ii) Kinetic energy, KE =GMem2r

(iii) Total energy, E=-GMem2Re

13. Time Period of Satellite:

T=2πr3GM=2πR+h3gR2

14. Height of Satellite:

h=T2gR24π213-R

15. Escape Speed:

The minimum velocity with which a body should be projected from the surface of the earth so that it escapes from the earth's gravitational field.

(i) ve=2GMR=2gR

(ii) ve=2×43πρGR×R ⇒ ve=R83πGρ

(iii) Escape speed from the surface of earth, ve=2gR=2×9.8×6.4×106≈11.2 km s-1

(iv) For nearby satellite, v0=GMR=ve2

(v) Escape speed does not depend on the angle of projection and the mass of the projected body. It depends on the mass and radius of the planet and the position from where the body is projected.

16. Kepler’s Laws of Planetary Motion:

(i) The law of orbits: Every planet moves around the sun in an elliptical orbit with the sun at one of the foci.

(ii) The law of area: The line joining the sun to the planet sweeps out equal areas in equal intervals of time. i.e. areal velocity is constant. According to this law, the planet will move slower when it is farther from sun and more rapidly when it is nearest to sun. It is based on the law of conservation of angular momentum.

What is the change in mass of a body when taken 64 km below the surface of the earth take radius of the earth as 6400 km?

Areal velocity, dAdt=12 rvdtdt=12rv

dAdt=L2m

(iii) The law of periods: The square of period of revolution T of any planet around the sun is directly proportional to the cube of the semi-major axis of the orbit.

T2∝a3 or T2∝r1+r223

17. Path of a Satellite with Different Speed of Projection:

What is the change in mass of a body when taken 64 km below the surface of the earth take radius of the earth as 6400 km?

Suppose a satellite is at a distance r from the centre of the earth. If we give different velocities v to the satellite, its path will be different.

(i) If v<vo, then the satellite will move is an elliptical path and strike the earth's surface. But if size of the earth were small, the satellite would complete the elliptical orbit, and the centre of the earth will be at is farther focus.

(ii) If v=vo, then the satellite will revolve in a circular orbit.

(iii) If ve>v>vo, then the satellite will revolve in an elliptical orbital and the centre of the earth will be at its nearer focus.

(iv) If v=ve, then the satellite will just escape with parabolic path.

(v) If v>ve, then the satellite will escape with hyperbolic path.

18. Geostationary Satellite:

The satellite which appears stationary relative to the earth is called a geostationary satellite. The orbit of a geostationary satellite is known as the parking orbit.

(i) It should revolve in an orbit concentric and coplanar with the equatorial plane.

(ii) Its sense of rotation should be the same as that of earth about its own axis i.e., in an anti-clockwise direction (from west to east).

(iii) Its period of revolution around the earth should be the same as that of earth about its own axis. ∴ T=24 h=86400 s

19. Weightlessness:

The weight of a body is the force with which it is attracted towards the centre of earth. When a body is stationary with respect to the earth, its weight equals the force of gravity. This weight of the body is known as its static or true weight. Weightlessness is the sensation of the absence of weight of a body. It is experienced in the following cases:

(i) When objects fall freely under gravity.

(ii) When a satellite revolves in its orbit around the earth.

(iii) When bodies are at null points in outer space. This is a called zero gravity region.