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PROBLEMS ON GRAVITATION

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PROBLEMS ON GRAVITATION

Q :1. You have system of two  point masses m1 and m2 , the distance  between them is r. they exert for F on each other. This system is on earth. Now if you take this system to moon,Find the force between them on moon.

(a) F/6

(b) F/12

(c) F/2

(d) None of these

Answer Option : (D)

Explanation : => The force between two point mass = Gm1m2 / r

=> This does not depends on the mass of earth or moon.

Q : 2. Time taken by the  planet to cover path A B C is t1 time taken by the planet to cover path C D A is t2.



(a) t1 = t2

(b) t2 > t1

(c)

(d) t1> t2

Answer Option : (D)

Explanation :
=>According to Keplersinglequotes second law ( law of areas);

dA / dT = Constant

=> Area covered by line during path A B C is more than the are covered by line during path CDA .

=> A1 > A2

=> So, t1> t2

Q : 3.



(a) Force on point mass m is more than  GmM /R2

(b) Potential (gravitational ) at point p = -GM / R

(c)
Force on point mass m is zero.

(d) Force on point mass is less than GmM / R2

Answer Option : (D)

Explanation :
The force on point mass m will be only due to sphere because it is outside the sphere and inside the shell (Sphere and shell are both have uniform mass density)


=> So Force on point mass = GmM / x2

(We assume  X is the distance of mass (point ) from center of sphere)
=>As X> R


So,



=> Potantial at point

Q : 4 . A particle is at distance 3000 km away from the surface of earth,The value of  g for the particle can be calculated  by the formula.

(a)

(b)

(c)

(d)

Answer Option : (B)

Explanation :
As, h = 3000 km
 
=> It is negligible and not very less than Re.

=> So we can not leave h in comparison to Re and also we can not use the formula



=>so the correct formula is  


Q : 5. Choose the correct option.

(a) A mass balance will  not tell the the correct mass of an object if taken on  moon .

(b) A spring balance will tell incorrect mass of an object if taken on moon.

(c) A mass balance will tell the correct mass of an object if taken on moon.

(d) a spring balance will tell correct mass of an object if taken on moon.

Answer Option : (C)

Explanation :






Now you can think why only mass balance will correctly on moon not spring balance.

Q :  6. We fill Weightlessness in satellite revolving around the earth.

(a) The net gravitational force on us in satellite is zero.

(b) Because there is centrifugal force which cancel the gravitational force.

(c) Because we are not able to touch the walls at satellite ,if we could just touch the we will get the normal force and hence we fell our weight.

(d) Net gravitational force on us is not Zero,But we Still Feel Weightlessness.

Answer Option : (D)

Explanation :


       

If satellite revolve around the earth in the circular motion, it requires a centripetal force ,which is provided by Gravitational force due to earth.

A person in the satellite is also revolving about the earth, so the gravitational force on the person is providing him/ her the required centripetal force.

=> Means gravitational force is acting on the person revolving on the satellite,but this gravitational force is acting as a centripetal   force . in this condition  if the person touch the floor (any surface) of satellite,It will not able to push that surface due to itsinglequotes weight and will not get the reaction.

=>We feel our weight when we get reaction from floor.So in this case we will fill weightlessness.

Q : 7. The P.E of satellite is -40 j and K.E is 20J.

(a) P.E < K.E

(b) P.E> K.E

(c) The minimum energy is required to throw the satellite out of earthsinglequotes gravitational field is +20 j

(d) The option (c) is correct because the K.E of the satellite is  +20 j so by this energy it can go outside the earthsinglequotes gravitational field.

Answer Option : (B),(C)

Explanation : P.E of the revolving satellite is greater than K.E of that revolving satellite.(Think why?)

=> The Minimum energy (external work done ) to throw the satellite out of earthsinglequotes gravitational field.

= -(T.E)

= -(-40+20)

= -(-20)

Wext = +20j

=> We know it we want to throw the satellite as it just cross the earthsinglequotes gravitational field,in this condition

=>Total mechanical energy (at the end of earthsinglequotes gravitational field) = K.E+ P.E

                                                                                                  = 0 + 0 = 0

=> charge of total mechanical energy = wext

= (T.E)f - (T.E)I

= 0 - (-20J)

Wext = + 20 J

So,



Q : 8. If the speed by which you throw a ball from earth surface is 2 ve , the speed vsinglequote when it just cross the gravitational field of earth..

(a)

(b)

(c)

(d)

Answer Option : (A) , (D)

Explanation :
If you use conservation of Mechanical energy.



Q : 9.
       
Find the height at which K.E of ball become zero.

(a) Re/2

(b) 2Re/3

(c) Re

(d) Out of earth gravitational field.

Answer Option : (C)

Explanation : From the conservation of mechanical energy.



Q : 10.


Assume at max height in the last question, some one catch that ball and how want to throw this ball out of earthsinglequotes gravitational field ,so what amount of min. speed he should give that ball how?
[ v is a speed of the ball on the earth surface in question ]

(a) v/2

(b) v/3

(c) 2v/5

(d)  v

Answer Option : (D)

Explanation :
Now for this question we can again apply conservation of mechanical energy.

Let the speed given to throw the ball out of earthsinglequotes gravitational field be vsinglequote at height hmax.



Same speed as it had on the surface of earth surface.


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