Gravitation - Free MCQ Practice Test with solutions, NEET Physics

To find the height above the Earth's surface where the gravitational acceleration is one-fourth of its value at the surface, we can use the formula for gravitational acceleration:

Thus, the height at which the gravitational acceleration is one-fourth of that at the Earth's surface is R_e. 

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

The weight of a body on Earth is the force due to gravity acting on it. This weight is calculated as the product of mass and gravitational acceleration.

• The Earth's rotation creates a centrifugal force that opposes gravity.
• If the Earth's rotational speed increases, this centrifugal force becomes stronger.
• As a result, the effective gravitational force on a body decreases, making the body weigh less.
• Therefore, with increased rotation, the weight of the body will decrease.

According to Newton's law of universal gravitation, the gravitational force (F) between two masses is inversely proportional to the square of the distance (r) between them, i.e., F ∝ 1/r². Therefore, when the distance is doubled, the force becomes one-fourth of the original force. 

d ∝ 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​−ω²R cosϕ
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

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

The total energy (E) of a satellite in orbit is the sum of its kinetic energy (KE) and potential energy (PE):

E = KE + PE

For a satellite in circular orbit:

If the kinetic energy is doubled, the new kinetic energy becomes:

The total energy becoming zero means the satellite has reached escape velocity. It will no longer remain in orbit and will escape into space.

The density of the planet remains the same, so the mass M of a planet is proportional to its volume, which depends on its radius.

The volume of a planet is proportional to R³, and since the density ρ is constant, the mass M is also proportional to R³.

Thus, we can write

Now, the escape velocity formula becomes:

This shows that the escape velocity is directly proportional to the radius of the planet. So, if the radius is doubled, the escape velocity will also double.

If the planet has twice the radius but the same density, the escape velocity will be 2v₀​.

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

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

C is correct.

For a particle of mass m at distance r from a central mass M, the total mechanical energy is E = ½ m v² - GMm/r.

If E = 0, then ½ m v² = GMm/r, which gives v = √(2GM/r), the escape velocity. A particle with this speed will reach infinity with zero kinetic energy, so it will escape the gravitational field.

Statement 1 is not always true. If the cavity is concentric with the sphere, the field inside the cavity is zero; if the cavity is off-centre, the field inside is uniform and non-zero. Therefore the claim that it must be non-zero and uniform is false.

Statement 2 is false at large heights where the central inverse-square law applies: trajectories under an inverse-square central force are conic sections. For horizontal speed less than the circular speed the path is part of an ellipse, not a parabola. A parabolic approximation applies only near the surface when the field is treated as uniform and Earth's curvature is negligible.

Statement 4 is false because satellites may have any orbital inclination; only a geostationary satellite must lie in the equatorial plane.

The solution to the problem can be explained as follows:

• The graph shows the relationship between energy and orbit radius for a body in circular planetary motion.
• There are three curves: A, B, and C, representing different types of energy.
• In this context:
  • Curve A represents kinetic energy.
  • Curve B represents potential energy.
  • Curve C represents total energy.
• The correct statement, according to the solution, is that A and B are kinetic and potential energies respectively, while C is the total energy.

To solve the problem of finding the increase in potential energy when a body of mass m rises to a height h = R/5 from the Earth's surface, we can follow these steps:
Step 1: Understand the Initial and Final Potential Energy
The potential energy P of a mass m at a distance r from the center of the Earth is given by the formula:

where G is the gravitational constant and M is the mass of the Earth.
At the surface of the Earth (where r = R ):

Step 2: Calculate the Final Potential Energy
When the body rises to a height h = R/5, the distance from the center of the Earth becomes:

Thus, the final potential energy is:

Step 3: Calculate the Change in Potential Energy
The change in potential energy ΔP is given by:
ΔP=P_final − P_initial 
Substituting the values we found:

Step 4: Relate to Acceleration due to Gravity
We know that the acceleration due to gravity g at the surface of the Earth is given by:

We can express GM in terms of g and R:
GM = gR² 
Step 5: Substitute G M into the Change in Potential Energy
Substituting G M into our expression for Δ P :

Step 6: Substitute R in terms of h
Since h = R/5, we can express R as:
R = 5h
Substituting this into the expression for Δ P:

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/r²)
Therefore g' = G{(1/10)M}/[{(1/3)r}²]
= (9/10) G(M/r²)
= (9/10)g
Now, H = {v²-u²}/2(-g)
= {0²- u²}/(-2g)
= {u²}/2g
H' = u²/[2{(9/10)g}]
= (10/9)u²/2g
= (10/9)H
= (10/9) × 90
= 100 m

Potential energy of the body at a distance 4Re from the surface of earth 

So minimum energy required to escape the body will be mgR_e/5.

When gravitational force on satellite suddenly disappears, the satellite will move with its velocity v tangent to its original orbit, due to inertia of motion.

Given,

r_A : r_B = 1.4 : 1

The total mechanical energy of the satellite will be

E = −G M m / 2r

E∝1 / r

ratio, E_B / E_A = r_A / r_B = 1.4 / 1

E_B : E_A = 1.4 : 1

Let x be the distance of the point of no net field from earth.
The distance of this point from moon is (r – x), where r = 3.8 × 10⁵ km.
The gravitational field due to earth = GM_e/x² and that due to moon = GMm/(r - x)². For the net field to be zero these are equal and opposite.

Acceleration due to gravity g = 
Thus = 1 ∴ g₂ = g₁ = g
Escape velocity V_e = 
∴ Escape velocity at planet = 
= (√2) (11.2km / sec ) = 15.84 km/sec.

The gravitational field at the given point is  

The mass M may be calculated as follows. Consider a concentric shell of radius r and thickness dr. Its volume is

and its mass is 
The mass of the whose sphere is 
Thus, by (i) the gravitational field is 

⇒ E = 418 m/s²

r =  minimum distance  
conservation of angular momentum about star mv_oL = mrv₁ 


Solving

Substituting the values of G, M, v₀​, and L: 

r = 3 × 10⁸ m = 3 unit

Work done by external agent
W_ent, = ΔU + ΔK 
= U_A – U_∞ + K_A - K_∞ 
= U_A - 10 + 1/2 × mv² - 0
-3 = U_A - 10 + 1/2 × 1 × 4
+ 5 J = U_A
So V_A = 5 J/kg