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Correct choice is (b) False

The BEST I can explain: A GEOSTATIONARY SATELLITE appears to be fixed in the sky because its time period of revolution around the EARTH is equal to the time period of revolution of the earth around its own axis, i.e., 24 hours.

The correct choice is (a) they cannot transmit data at long distances DUE to curvature of the earth

Easy EXPLANATION: Most communication waves cannot transmit data accurately over long distances due to the curvature of the earth. They can only do so in the line of SIGHT. However, they are not reflected by the ionosphere like radio waves and can be relayed to a wide AREA using geostationary SATELLITES.

Right choice is (b) 4:3

The explanation: ORBITAL velocity is inversely proportional to the square ROOT of the orbital radius.

Hence;

v1 : v2 = 16^1/2 : 9^1/2

 = 4 : 3

The correct choice is (B) of VISCOUS forces causing the speed of the satellite and HENCE height to gradually decrease

The best explanation: DUE to the presence of some gases of the atmosphere and other debris, the satellites speed is reduced due to viscous forces. This reduces the height of the satellite. Hence, it eventually falls to the earth.

Right option is (C) 25 J

Explanation: The gravitational potential energy, in vectorial mathematics, is the dot product of the gravitational field and relative position MULTIPLIED by mass.

Gravitational potential energy = (g . P) X m

 = [(2i + 3j) .(10I – 5J)] x 5

 = (20 – 15) x 5

 = 25 J.

Correct CHOICE is (a) [(n*G*M)/(n+1)R] ^1/2

For explanation I WOULD say: Total INITIAL energy = Total final energy

1/2(m X v^2) – (G*M*m)/R = -(G*M*m)/(n+1)R

Simplifying the above equation, we get;

v = [(n*G*M)/(n+1)R]^1/2.

Right option is (b) V/4r

Best EXPLANATION: Gravitational potential ENERGY (V) at a distance “r”;

V = -(G*M*m)/r

Gravitational force at a distance “r” = -(G*M*m)/r^2

 = V/r

Gravitational force at a distance “2R” = -(G*M*m)/4r^2

 = V/4r.

Right option is (b) False

Easy explanation: PSLV stands for “POLAR Satellite Launch Vehicle”. They launch polar SATELLITES which are low-earth orbit satellites. They orbit the earth at about 500 – 800 km from the earth’s surface.

The correct choice is (a) True

The explanation is: The ENTIRE planet EARTH has only 1 natural satellite; the moon. All other satellites are CALLED “artificial satellites”. India has launched approximately 118 artificial satellites since 1975.

Right ANSWER is (a) True

The explanation is: For a satellite of mass “m” revolving around a PLANET of mass “M” and RADIUS

“R” at a height “H” above the surface, the satellite traverses a DISTANCE [2 x pi x (R + h)] in time “T” if the orbital velocity is “v”.

Hence, the time periods (T) = [2 x pi x (R + h)]/v

We know; orbital velocity (v) = [(G x M)/(R+h)]^1/2

Hence; T = [2 x pi x (R + h)^3/2/(G X M)^½], which is independent of the mass of the satellite.

The correct option is (a) the mean VELOCITIES of ATMOSPHERIC gas MOLECULES which is less than that of the earth’s escape velocity

For explanation: The escape velocity of earth is approximately 11.2km/s. This is greater than the root-mean-square velocities of the atmospheric GASES. HENCE, the gases are unable to escape into the vacuum of space.

The correct option is (a) True

Easy explanation: The GRAVITATIONAL potential INSIDE a uniform thin spherical shell of radius “a” is;

V = -(G*M)/a, which is a constant. Hence, the value of gravitational potential inside a uniform thinspherical shell is the same everywhere and not necessarily non-zero.

Correct CHOICE is (a) 24 x (3/5)^1/2 hr

Easiest explanation: The time PERIOD of a satellite is given by;

T = 2 x pi x (R+h)^3/2 / (G x M)^1/2

For a geostationary satellite;

24 = (5R)^3/2 x (constant); [As per GIVER information]

Required;

t = (3R)^2/2 x (constant)

t = Time period of the satellite at a height of 2.5R

t/24 = (3R/5R)^3/2

 = (3/5)^3/2

t = 24 x (3/5)^1/2 hr.

The CORRECT option is (a) True

To elaborate: A polar orbit is IDEAL for a spy satellite because it is CLOSER to the earth’s surface and can CLICK high-resolution images of the surface and also pick up communication data that is being transmitted.

The correct option is (b) decrease 10 times

Explanation: New mass of the EARTH (M’) = 0.01 X ORIGINAL mass of the earth (M)

Therefore;

M’ = M/100

Escape velocity (v) is directly proportional to the square root of the mass

“v” is directly proportional to the square root of “M’”.

Therefore;

New escape velocity (v’) = v/10.

The correct ANSWER is (d) 64:1

The best I can EXPLAIN: The time periods (T) of a SATELLITE revolving around the earth of radius R and at height h is:

T = 2 x pi x [(R + h)^3 / (G X M)]^1/2

Hence, the time period is directly proportional to (R + h)^3/2

The ratio of ORBITS is, therefore, 64:1.

Correct CHOICE is (b) 12.15 x 10^30 kg

Best explanation: ESCAPE velocity (v) = [(2 x G x M)/R] ^1/2

M = Mass of the black hole

R = Radius of the black hole

3 x 10^8 = [(2 x 6.67 x 10^-11 x M)/18000]^1/2

9 x 10^16 = (2 x 6.67 x 10^-11 x M)/18000

M = 12.15 x 10^30kg.

Correct choice is (a) 100 J

Easy EXPLANATION: The work done to raise the BIRD’s own centre of mass to the BRANCH depends only on the mass of the bird.

Hence, the work done = (4 x 10 x 2.5) J

 = 100 J.

Right option is (b) 1:2

The explanation is: Let the MASS of the SATELLITE be “m” and mass of the planet be “M”.

KINETIC energy (KE) = 1/2 x m x v^2

 = (1/2) x (G x m x M)/R

Potential energy (PE) = – (G x m x M)/R

KE : PE = -1:2

Therefore, (the magnitude of KE:PE) = 1:2.

The correct option is (a) -(G*M*m)/12R

For explanation I would SAY: Total ENERGY at 1.5R (TE) = -[(G*M*m)/3R]

Total energy at 2R (TE’) = -[(G*M*m)/4R]

The difference in energy is the REQUIRED energy.

Difference = TE’ – TE

 = -[(G*M*m)/4R] – [-[(G*M*m)/3R]]

 = -[(G*M*m)/4R] + (G*M*m)/3R]

 = (G*M*m)/12R.

Correct OPTION is (a) True

Best explanation: The orbital RADIUS “r” for a geostationary satellite is as follows;

r = [(T^2 x G x M)/(4 x pi^2)] ^1/3

 = [(86400^2 x (6.67 x 10^-11) x (6 x 10^24))/(4 x pi^2)] ^1/3

 = 4,22,00,000 m

 = 42,200 km

r = R + h

R = Radius of earth

 = 6,400 km

h = height of satellite above the earth’s surface

42,200 = 6,400 + h

h = 35,800 km

 = APPROX. 36,000 km.

The correct ANSWER is (d) (r*V^2/R^2)

Explanation: Orbital VELOCITY (v) = [(G x M) / r]^1/2

(G x M) = v^2 x r

Acceleration DUE to GRAVITY (g) = (G x M) / R^2

Therefore; g = (v^2 x r) / R^2.

The correct option is (C) 64(W^2 X (nQ) ^3/Q^2)

For EXPLANATION: Orbital velocity (v) = [(G x M) / (4nQ)]^1/2

(G x M) = v^2 x (4nQ)

Acceleration DUE to gravity (g) = (G x M) / Q^2

Therefore; g = (v^2 x (4nQ)) / Q^2

Angular velocity (w) = v/(4nQ)

v = w x (4nQ)

Therefore; g = (w^2 x (4nQ)^3/Q^2)

 = 64 x (w^2 x (nQ)^3/Q^2).

The CORRECT option is (d) [(2*G*M)/R]^1/2

To explain: If the object’s initial velocity is “v”, from the LAW of conservation of ENERGY, we have;

(m X v^2)/2 = (G x M x m)/R

v^2 = (2 x G x M)/R

Therefore; v = [(2 x G x M)/R]^1/2.

Right ANSWER is (c) PI/6

The BEST I can EXPLAIN: Angular velocity of the satellite if the EARTH were stationary; w’ = (2 x pi) / 8

 = pi / 4

Angular velocity of earth; w’’ = (2 x pi) / 24

 = pi / 12

Therefore, the angular velocity about the centre (w) = w’ – w’’

 = pi/4 – pi/12

 = pi/6

The correct choice is (a) True

The explanation is: Since the reference for gravitational POTENTIAL energy is taken at INFINITY, conventionally, we DENOTE any finite value of gravitational potential energy, except zero, as a negative number. Hence, the greatest value of gravitational potential energy is zero.

The correct option is (B) 1.875 J

The explanation is: The work done by the eagle on the SNAKE DEPENDS only on the MASS of the snake.

Hence, the work done on the snake = (0.075 x 10 x 2.5) J

 = 1.875 J.

Right ANSWER is (a) True

Explanation: The escape velocity for any object on EARTH is approximately constant at all points on the earth’s surface, i.e., the escape velocity of earth is approximately equal to 11.2 km/s and is not determined by any PARAMETER of the PROJECTILE.

Right option is (b) False

For explanation: Gravitational POTENTIAL at a point can be DEFINED as the WORK done in bringing a UNIT mass from infinity to that point.Hence, for a given point mass at infinite DISTANCE, both, the gravitational potential and the gravitational potential energy can be zero.

Right answer is (c) directly PROPORTIONAL to P^1/2

Best explanation: Escape velocity (v) = [(2*G*M)/R] ^1/2

Density (P) = Mass (M) x Volume

M = P / [(4/3) x pi x R^3]

THEREFORE;

v = [(8 x pi x P x G x R^2) / 3]^1/2.

Right choice is (b) -(G*M)/a

Explanation: We KNOW that the gravitational POTENTIAL at a point “P” at a distance of “r” from the centre of the RING of radius “a” and mass M” is;

V = -(G*M)/(a^2+r^2)^1/2

At the centre of the ring; r = 0.

Therefore; V = -(G*M)/a.

Correct choice is (d) [(5*G*M)/(6*R)]^1/2

Explanation: The orbital velocity of a satellite flying at a HEIGHT “H” above the surface of the earth of radius “R” is:

V = [(G*M)/(R+h)]^1/2

Here, h = R/5

Therefore; v = [(5*G*M)/(6*R)]^1/2

Right choice is (b) False

For explanation I would SAY: The gravitational potential outside a thin uniform spherical shell behaves similar to the potential DUE to a POINT source FORM the centre of the spherical shell.The gravitational potential outside a spherical shell is inversely proportional to the distance of the point outside from the centre of the sphere. Hence, it is not the same EVERYWHERE.

Correct option is (a) True

To elaborate: The potential energy of a satellite is;

PE = – (G X m x M)/(R+h)

h: Height above the surface of the planet

For an elliptical orbit, “h” changes with time. HENCE, the potential energy ALSO varies with time.

Correct answer is (C) 90 degrees

Best EXPLANATION: From the figure, we can SEE that the POLAR orbit has an inclination of 90 degrees from the equatorial PLANE.

The correct ANSWER is (a) – (m*v^2)/2

Easy EXPLANATION: TOTAL energy (E) = KE + PE

KE = (1/2) x m x v^2

PE = – 2 x (KE) [For a satellite]

 = – m x v^2

E = (1/2) x m x v^2 + (- m x v^2)

 = – (1/2) x m x v^2.

Correct option is (B) radius of its orbit

Easy explanation: The time period of a satellite is GIVEN by;

T = 2 x pi x [(R + H)^3 / (G X M)]^1/2

Where;

R = Radius of planet

h = Height above the planet

M = MASS of the planet

(R+h) = Radius of orbit

The correct choice is (a) (pr)^1/2

The best EXPLANATION: Escape velocity (v) = (2*g*R)^1/2

g = ACCELERATION due to gravity

R = Radius of planet

Given;

 Ra/Rb = r

 ga/gb = P

va/vb = (2*ga*Ra])^1/2/(2*gb*Rb) ^1/2

 = (p X r)^1/2.