Test: Gravitation - JEE MCQ

A communications satellite is in a GEO (geostationary equatorial orbit) with a period of 24 hours. The spin rate ω_s about its axis of symmetry is 1 revolution per minute, and the moment of inertia about the spin axis is 550 kg · m².

g = GM​/R² 
g at height h
g_h​ = GM/(R+H)²
h = R/2
g_h ​= GM/(R + R/2​)² ​= GM/((9/4) (R²)) ​= 4/9g
Decrease in the value of g
g − 4/9g = 5/9 g

Force on the body placed on Earth's surface is
F = GMm/R²​
But, F=mg hence,
mg = GMm​/R²
where, variables have their usual meanings.
gR² = GM
Now, force on the body at geo-potential height say h (altitude) where the acceleration due to gravity is 25% of that at the earth's surface i.e.
25//100g ​= g/4​
Hence, we can write
g/4​ = GM/(R+h)²​

g/4​ = gR²/(R+h)²​

(R+h)² = 4R²

Taking roots for both sides we get
R+h = 2R
h = R

If M is the mass of the Earth and m is the mass of a sphere of radius r and you want the force of gravity to be the same at both the surface of the sphere and the surface of the earth then mg = Mg and m = M. This means that the mass of the sphere would have to be the same as the mass of the Earth no matter what r is. This is impossible since a sub-sphere cannot have the same mass as its parent.
For a sphere of radius r to have force g at its surface, d = (De *Re)/r where De is the density of the entire Earth and Re is the radius of the Earth, d is the density of the sphere of radius r. So there is a radius and a density which will produce a gravitational force of g for any d, r >0 but one cannot be a subset of another.

= d μ 1/r

Let,  g_e​ be the acceleration due to gravity on surface of earth, g′ be  the acceleration due to gravity at depth d below the surface of earth and is given by
g′=g_e​(1−d/R​) ............................(1)
The acceleration due to gravity of an object on latitude(below the surface of earth) is
g′=g_e​−ω²Rcosϕ
where, ϕ is the solid angle at the centre of earth sustained by the object.
At the equator, ϕ=0∘
∴cosϕ=1
Hence, 
g′=g_e​−ω²R ...............(2)
From (1) and (2), we can write
g_e​(1−d/R​)=g_e​−ω²R 
⇒dg_e​​=ω²R
⇒d= ​ω²R²​/ g_e

The acceleration due to gravity on surface of earth is
g=GM​/ R²
 
g=GρV​/R²
where, G is the gravitational constant, M  is mass of earth, R  is radius of earth, ρ is the density and V is the volume of earth.
∴g=Gρ(4/3)​πR^3​/ R²
 
∴g=(4/3)​GρπR
⇒g∝ρR
since, all other quantities are constant.
Hence, if the radius of the earth be increased by a factor of 5, density will decrease by 5  to keep the value of g the same.

As M_p​=2M_e​ and D_p​=2D_e​
or R_p​=2R_e​
Hence, g_p​= GM_p/(R_p​)² ​​=G(2M_e​)/(2R_e​)2 ​= GM_e/2R_e²​ ​​=g_e/2​​
Time period of pendulum on the planet  T_p​=2π√ l/gp​
Tp​=2π√​2l/g_e​​=√2​×2π√l/g_e​=√2​×T_e​
T_p​=√2​×2=2√2​s

According to the question, there can be two cases in which the point mass is at a distance a from the surface of the spherical shell. 

1. The point mass is at the center of the shell

2. The point mass is outside the shell at a distance 2a from  the center of the shell.

Superposition principle dictates that the potential and field at any point will be the sum of the potentials and fields respectively at that point due to the individual masses.

It can be seen that in both the cases, neither field nor potential is zero inside the shell. This is due to the contribution of the point mass.

Let the density be d for both the planets. Given that R_A​=2R_B​
Now, mass of A, M_A​=4dπR_A​³/3​=32dπR_B​³​/3
similarly, M_B​=4dπR_B​³/3
Escape velocity for a planet is given by V=√2GM​​/R
So, V_A​=​√2GM_A/3R_A​​​=√​64GdπR_B​³/6R_B ​​=√32GdπR_B​²/3​​
 
Similarly, V_B​=8GdπR_B​²/3​​
 
Taking the ratio, ​V_A/V_B ​​=32GdπR_B​²​/3​×​√3/8GdπR_B^​2​​=2

Gravitational Potential V = -GM/R for hollow spherical shell at the centre. If we replace R by R/2 then we get V = -2GM/R. Therefore it decreases.

g∝1​/R²
Curve b represents this variation.
 

Here the satellite is moving with constant velocity v and it is the gravitational force which is the necessary centripetal force to allow it to move around the earth.
Imagine this as: you are holding a string attached to a stone at other end and you whorl it in the air, now when you suddenly leave the string the stone will move outwards. This is actually tangential to the direction of previously applied force.
When this same concept is applied to the satellite, it will also move away tangentially and with uniform velocity due to zero net force in outer space and no air resistance (as space has vacuum).

Let the velocity of m is v and velocity of 4m is u
Therefore, v = 4u and we know that kinetic energy is 1/2mv² so for m:
KE = ½ x m x v² = 1
And for 4m:
KE = ½ x 4m x 4² = 4
So, ratio of their kinetic energy is 1:4
Hence A

If the TE = 0 that means KE + PE = 0
KE = - PE which is enough for the body to leave the system (here, Earth)

The body is in circular planetary motion this implies that Potential energy will be negative (attractive)  and the total energy of the system will lie between zero and potential energy (because of circular motion).

Both the objects with different masses will attain the same v when striking the ground because acceleration and height is the same. 
As, K.E=(1​/2)mv²P.E=−GMm​p/r=mv
Thus K.E, P.E, and p will be different.
 

Initial PE = -GMm/R
Final PE = -GMm/R+h, (h=R/5)
Increase in PE = final - initial
= -GMm/R + (R/5) - (-GMm/R)
=GMm(1/R - 5/6R)
=GMm/R * 1/6
= gmR/6 (g= GM/R)
=gm5h/6 (h=R/5)

Let mass of earth be M
Acceleration due to gravity = g
Radius of earth = r
Height to which he can throw on earth = H
Acceleration due to gravity at the planet be = g'
Radius of the planet be= r'
Mass of the planet be = M'
Height to which he can throw at that planet = H'
Then g = G(M/r^2)
Therefore g' = G{(1/10)M}/[{(1/3)r}^2]
= (9/10) G(M/r^2)
= (9/10)g
Now, H = {(v)^2-(u)^2}/2(-g)
= {(0)^2-(u)^2}/(-2g)
= {(u)^2}/2g
H' = u^2/[2{(9/10)g}]
= (10/9)u^2/2g
= (10/9)H
= (10/9) * 90
= 100 m

Potential energy at surface of earth will be PE₁ = (G.M.m)/R

Now, potential energy at height nR from the surface of earth,

PE₂ = (G.M.m)/(R+nR)

Here, change in PE = PE₁ – PE₂

= (G.M.m)/R – (G.M.m)/(n+1)R

= (G.M.m)/R[1 – 1/(n+1)]

= (G.M.m)/R[(n + 1 – 1)/(n+1)]

= [n/(n+1)](G.M.m)/R ….(i)

Also, g = GM/(R^2) ….(ii)

From (i) and (ii),

Change in PE = (n/(n+1))mgR

Obviously, take example that u have one thread in your hand with stone tied at one end and your moving it in vertical circular manner and if u suddenly leave the thread it will go in the direction tangentially to your arm here your hand force is equal to gravitation force but your hand plays centrifugal and gravitation plays centripetal.

We know that,
Hence the ratio of total energy is in the inverse ratio of the distance ‘r’.