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**Q.1. A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches height R(R = radius of the earth), it ejects a rocket of mass m/10 so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G is the gravitational constant; M is the mass of the earth) (2020)(1)(2)(3)(4)Ans. **(2)

Solution.

According to law of conservation of energy, we have

Initial energy = Final energy

K

Where K

................. (1)

Using conservation of momentum, we have

From Eq. (1), we get

So, kinetic energy of rocket is given by

**Q.2. A box weighs 196 N on a spring balance at the north pole. Its weight recorded on the same balance if it is shifted to the equator is close to (Take g = 10 m/s ^{2} at the north pole and the radius of the earth = 6400 km) (2020)(1) 195.66 N(2) 194.32 N(3) 194.66 N(4) 195.32 NAns. **(4)

Solution.

Given that, mg = 196 N ⇒ m = 19.6 kg

R = 6400 km = 6400 × 10

We know that, weight at equator is given by W= mg - mω

(1) 2/3

(2) 1/6

(3) 1/2

(4) 1/3

Gravitational field on the surface of a solid sphere is given by

For Sphere 1, we have

.............. (1)

For sphere 2, we have

................. (2)

From Eqs. (1) and (2), we have**Q.4. An asteroid is moving directly towards the centre of the earth. When at a distance of 10R (R is the radius of the earth) from the earths centre it has a speed of 12 km/s. Neglecting the effect of earths atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is 11.2 km/s). Give your answer to the nearest integer in kilometer/s _______. (2020)Ans. **(16)

Solution.

Since, there is no loss of energy. Then, conservation of energy we have

Total energy of asteroid at distance 10R = Total energy of asteroid at distance R

K

Given that ν_{e }= 11.2 km/s, v_{i} = 12 km/s

So, **Q.5. A body A of mass m is moving in a circular orbit of radius R about a planet. Another body B of mass m/2 collides with A with a velocity which is halfof the instantaneous velocityof A. The collision is completely inelastic. Then, the combined body(1) Continues to move in a circular orbit(2) Escapes from the Planet’s Gravitational field(3) Falls vertically downwards towards the planet(4) Starts moving in the elliptical orbit around the planet (2020)Ans. **(4)

Solution.

Given that the body of mass m is moving in a circular orbit of radius R.

So, its orbital speed is given by

Where M is the mass of the planet.

Since, collision is inelastic. So, from conservation of linear momentum we have

Since after collision, the speed is not equal to orbital speed at that point. So, motion cannot be circular. Since velocity will remain tangential, so it cannot fall vertically towards the planet. Escape velocity of the object from planet is given by

So, ν_{f}_{ }< ν_{e}

Their speed after collision is less than escape speed, so they cannot escape gravitational field. So their motion will be elliptical around the planet.**Q.6. Planet A has mass M and radius R. Planet B has half the mass and half the radius of Planet A. If the escape velocities from the Planets A and B are v _{A} and v_{B}, respectively, then v_{A/}v_{B }= n/4. The value of n is (2020)**(1)

(1) 4

(2) 1

(3) 2

(4) 3

Ans.

Solution.

Escape velocity from the planet is given by

Given that

So,

We have

Therefore, n/4 =1 ⇒ n = 4

(1) L /m

(2) 4L/m

(3) L /2m

(4) 2L/m

Ans:

Consider a planet moving in an elliptical orbit from point A to B and traces small area dA at the focus in time dt. Let the angle traced by radius vector be . The area of sector is given by

Now the instantaneous angular momentum is given by

Substitute the value in Eq. (1), we have

(1) 1.6 × 10

(2) 3.2 × 10

(3) 6.4 × 10

(4) 1.28 × 10

Ans:

The energy required to take satellite at height h above earth surface is equal to change in potential energy, that is, E

If satellite revolve around the earth at a distance (R

Therefore, E

**Q 9. A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass m is ejected from the satellite such that it just escapes from the gravitational pull of the Earth. At the time of ejection, the kinetic energy of the object is (2019)(1) 2mv ^{2}(2) mv^{2}**

(4) 3/2 mv

Ans:

By conservation of energy, we have

KE of Particle having mass

(PE = KE = 0)

KE of particle having mass

Therefore, KE of particle having mass m = mv

(1) 2.4 × 10

(2) 1.4 × 10

(3) 3.8 × 10

(4) 2.8 × 10

Let v be the minimum speed of no meteorite

(1)

(2)

(3)

(4)

Ans:

We have,

Centripetal force = Gravitational force

Now, change in velocity = v

(1)

(2)

(3)

(4)

Ans:

The mass per unit length of the rod,

dm = (A + Bx

Gravitational Force is

Integrating both the sides, we get

(1) such that it escapes to infinity.

(2) in an elliptical orbit.

(3) in a circular orbit of a different radius.

(4) in the same circular orbit of radius R.

Ans:

Along x axis by conservation of momentum

Along y axis by conservation of momentum

Total velocity

So, it is an elliptical orbit.

(1) ½

(2) 1

(3) 2

(4)

Ans:

Kinetic energy

We know that orbital velocity is

Now, kinetic energy for satellite A

Therefore, ratio of KE is

(1)

**(2) **

**(3) **

**(4) Ans:** (3)

**Q 16. Take the mean distance of the moon and the sun from the earth to be 0.4 x 10 ^{6} km and 150 x 10^{6} km respectively. Their masses are 8 x 10^{22} kg and 2 x 10^{30} kg respectively. The radius of the earth is 6400 km. Let ΔF_{1} be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and ΔF_{2} be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to ΔF_{1}/ΔF_{2} is: (2018)** B

(1) 10^{−2}

(2) 2

(3) 6

(4) 0.6

Ans:

= (375)^{3} (4 × 10^{−8})

= (0.37 × 10^{3})^{3}(4 × 10^{−8})

= 1.64**Q 17. The relative uncertainty in the period of a satellite orbiting around the earth is 10 ^{–2}. If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is: (2018)(1) 2 × 10^{–2}(2) 6 × 10^{–2}(3) 3 × 10^{–2}(4) 10^{–2}Ans:** A

From kepler's Law

(1)

(2)

(3)

(4)

Ans: (2)

Solution:

Variation of g inside earth surface

(1) 0.63 × 10

(2) 0.28 × 10

(3) 1.1 × 10

(4) 0.83 × 10

Ans:

Ans:

**(2) **

**(3) **

**(4) Ans:** D

Orbital velocity v =

Velocity required to escape

∴ Increase in velocity

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