Conceptual Question and Numerical Problem from Gravity and Gravition

Question 1: If earth suddenly stops rotating about its axis. What would be the effect of g? Would this effect be same at all places?

Answer: 

The variation of acceleration due to gravity (g) due to rotation of earth is:

g1=g-rw^2cos^2Ɵ

Then, w becomes zero (i.e. w=0) 

Now, g1=g-R*0^2cos^2Ɵ

i.e. g1=g (Maximum) 

The value of g can be changes place to place i.e. at equator, its value is minimum and at pole, its value is maximum.

Question 2: At what condition does a body becomes weightless at equator?

Answer:

Acceleration due to gravity at equator is:

g1=g-Rw^2cos^2Ɵ [ Ɵ=0]

Or, g1=g-Rw^2 (cos0)^2

i.e. g1=g-Rw^2

for weightless,

g1=0

g=Rw^2

i.e. w=√g/R

Question 3: Suppose a body is carried from earth to moon, what happen to its weight?

Answer:

We know that as we go , higher value of g decreases and it will be zero in the space (The region where gravitational force due to earth is equal to gravitational force due to moon). But, as we go further up, acceleration due to moon will be active. The value of it will be maximum on the moon surface.

Question 4: Why is G called Gravitational constant? Or What are its properties?

Answer:

G is called universal gravitational constant because:

1) G is independent of nature of medium.

2) It is independent of temperature.

3) It is independent of nature of masses.

Question 5: Find the acceleration due to gravity on the earth surface and hence calculate period of revolution of satellite in a circular orbit round the earth at height  of 306*10^6m above the earth surface?(Radius of earth =6.4*10^6m, masses of earth=6*10^24kg, G=6067*10^-11nm^2/kg^2)

 solution:

 we have g=Gm/R^2

                 =9.81ms^-2    

 Again, T= (2∏(R+h)^3/2)/√GM

           =9.911*10^3sec 

Question 6: Explain Newton's law of gravitation and Acceleration due to gravity on earth surface.

Answer:

Newton law of gravitation:

According to newton's law of gravitation, "  Every body in the universe attracts every another body with a force which is directly proportional to their masses and inversely to the square of distance between their center ".

i.e. F  α m1*m2...equation (1)

Fα 1/R^2 ....equation (2)

combing on above relation,

Fαm1*m2/R^2

F=Gm1*m2/R^2 where G is called gravitational universal constant, where its value is  6.67*10^-11. Therefore, the force of attraction between the two bodies in the universe is known as force of gravitation and force of attraction of acting on any other body by the earth is called gravity. 

Specially, G can be defined as:

M1=M2=1units, R=1unit then,

F=G

hence ,

The force of attraction between two bodies is of unit masses placed at unit distance.

 

Acceleration due to gravity on the earth surface:

when a body is released from a height, it falls downward and  it's velocity goes on increasing this mean that there is  a change in velocity at every time. The change in velocity per second is the acceleration.Hence, the acceleration produced in a body  in its free fall towards the earth is called acceleration due to gravity(g).              

Let 'R' and  'M' be radius and masses of earth  (m) be the masses of body placed on the earth surface. Then force of attraction between earth and body is:

F=GM*m/R^2

Or, mg=GM*m/R^2

Or,g=GM/R^2

i.e. g=GM/R^2

Explain Orbital velocity and Time period of earth surface.  

Answer:

 

Orbital velocity and time period of a satellite :

The orbital velocity of a satellite is the velocity required to be given to the satellite in order to put it in an orbit around the earth. For continuous revolution of satellite around of earth, centripetal force is equal to gravitational force developed on a satellite.

Let m and R be masses and radius of earth .If M be the masses of satellite  moving in the  orbit of radius (r) of height (h) from the surface of earth (i.e. r=R+h)

Then,

Fc= gravitational force

or, mv^2/R=GMm/R^2

V=√GM/R

V=√GM/R+h....equation (1)

If h=0 then

V=√GM/R

Again,

GM=gR^2....equation (2)

Then above equation (1) becomes 

V=√gR^2/R+h

And equation (2) becomes

V=√gR

Time period (T):

Here ,

T=2∏/w

T=(2 ∏)/v/r

T= 2∏r/v                         

T=2∏r/ (√gR^2/R+h)

T=2∏(R+h) √(R+h)/gR^2

T={2∏(R+h)^3/2}/(gR^2)^1/2

T= 2 ∏√(R+h)^3/gR^2

which is the time required solution.