Test: Gravitation - 2 - NEET MCQ

The resultant of these two forces is  .Now 

Force on the body placed on Earth's surface is
F=GMm/R^2​
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.
25g/100​=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
 

Given λ=(A+Bx²),
Taking small element dm of length dx at a distance x
from x = 0

So, dm = λ dx
dm = (A+Bx²)dx
dF = Gmdm / x²

g' = acceleration due to gravity at latitude
g = acceleration due to gravity at poles
ω = angular velocity of earth
(θ) = latitude angle
Now,
g' = g - Rω² cos²(θ)
As, g' = 0 (weightless):
g = Rω² cos²(θ)
g = Rω² cos²(30)
g = Rω² (3/4)
ω² = 4g/3R
ω = √(4g/3R)

Here u i = gravitational potential energy at the centre of earth

u = gravitational potential energy at the surface of earth

When the person started flying towards the planet, the person experiences the gravitational force due to the gravity of the planet, the force experienced by the man is 

When the person heads inside the planet through the hole, the gravitational force that he was once experiencing due to the shell of the planet is now gone and due to this the gravitational force on him becomes zero.
Now, when the person enters through the hole, he sees two shells of mass 2M/3 and point mass M/3. Thus the gravitational force acting on the person due to the point of mass of the shell is given as:

Now that we have gained the two energies that are one outside the planet and another inside the planets due to another shell-like structure, the net gravitational force acting on the person is the value of the second shell gravitational force subtracted from the first shell gravitational force.

  (orbital speed υ₀ of a satellite)

Near the earth’s surface is equal to 1/√2
times the escape velocity of a particle on earth’s surface) Now from conservation of mechanical energy: Decrease in kinetic energy = increase in potential energy

We use the energy balance
Initial potential energy equals final kinetic energy
Gm₁​m₂/d​​=1mv₁²​/2+1​mv₂²​/2
also, from the conservation of momentum we have
m₁​v₁​=m_2​v₂​
or
v_1​= ​m₁​v₁/ m₂
Substituting this we get 
v₁​=√2Gm₂²​​​/d(m₁​+m₂​)
Similarly, we have 
v_2​= √2Gm₁^2​​​/ d(m_1​+m₂​)
Now as velocities are in opposite direction their relative velocity is v₁​−(−v₂​)=v₁​+v₂​
or
[2G(m₁​+m_2​)​​/ d]^1/2

Energy required to raise a satellite upto a height h:

We know that attraction at an external point due to spherical shell of mass M is GM/r² while at an internal point is zero. So, for particle at point C
OA = G (M₁ + M₂)m/p²
OB = G M₁m/q²
OC = 0
 

By the principle of superposition of fields

Here  = net field at the centre of hole due to entire mass

 = field due to remaining mass

 = field due to mass in hole = 0

Let the acceleration due to gravity at that height is g′.

The acceleration due to gravity at some height (g′ ) = 1% of the acceleration due to gravity(g ) at the surface of the earth.

According to question

g′ = 1% of g

g′ = 1 / 100 × g

g′ = g / 100

The acceleration due to gravity at a height h above the surface of the earth is;

g′ = g (R / R+h)² , Where R is the radius of the earth.

Thus,

g / 100 = g(R / R+h)²

1 / 100 = (R / R+h)²

Taking square root on both sides we get,

1 / 10 = (R / R+h)

R+h = 10R

h = 9R

W = increase in potential energy of system

=U_f - U_i

=m (V_f - V_i)

(V = gravitational potential)

Note: Even if mass is nonuniformly distributed potential at centre would be -GM/R

Using formula for acceleration due to gravity,

When the radius is increased to five times,

Here mass of the earth is,

In order to maintain the same value for acceleration due to gravity:
g=g′

The center of mass (CM) for two point masses can be calculated using the formula:

Here, let m₁ = 4m (mass at position 0) and m₂ = m (mass at position d).

Now, we can find the distances of each mass from the center of mass:

For circular motion, the centripetal force acting on each mass is given by:

F_c = m⋅r⋅ω²

Where r is the distance from the center of mass and ω is the angular velocity.

For mass 4m:
F_c,4m = 4m⋅d / 5⋅ω²

For mass m:
F_c,m = m⋅4d / 5⋅ω²

Since both masses are in circular motion due to their mutual gravitational attraction, we can set the centripetal forces equal to each other:

4m⋅d / 5⋅ω² = m⋅4d / 5⋅ω²

The kinetic energy (KE) for each mass in rotational motion is given by:

K.E. = 1 / 2Iω²

Where I is the moment of inertia. The moment of inertia for point masses is given by I = m⋅r².

For mass 4m:

For mass m:

Now, we can find the ratio of the kinetic energies:

Thus, the ratio of the kinetic energies of the two masses is:

Ratio of Kinetic Energies = 1 : 4

The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just barely provides the centripetal force needed for the rotation. Let M be the mass of the planet R its radius and m the mass of a particle on its surface. Then

Net force is towards centre

Angular momentum of planet about the sun is constant.

i.e. mυr sin θ = constant

At position P and Q, θ = 90⁰ and m = mass of planet = constant

∴ υr = constant

From conservation of mechanical energy

in horizontal direction  

in vertical direction

 g=GM/r2                                                g1=GMr/R3
we are given that,
GMr/R³ =(1/4) GM/(R+R)²
=>r/R³=(1/4).(1/4R²)
=>r=R/6
Therefore, d=R-r=15R/16

Since we know that and ATQ T_A = 8T_B so,
⇒ (Ta/Tb)² = (Ra/Rb)³
⇒ 64 = x³
⇒ x = 4

on solving we get;h = r

Gravitational field due to the ring  at a distance d = √3R on its axis is:

force on sphere