Explore topic-wise InterviewSolutions in .

The ability of a body (planet) to hold the atmosphere depends on 

(i) acceleration due to gravity, and 

(ii) surface temperature.

At the poles

I will weigh more at poles and weigh less at the equator since weight is the amount of force with which an object is pulled towards the center of earth.

W = m × g ….1

Where, W = Weight

m = mass

g = Acceleration due to gravity

Since, mass is constant all the time, so,

W ∝ g ….2

And we also know that,

g = G\(\frac{M}{R^2}\)

Where, G = Gravitational Constant

R = Radius of earth

M = Mass of earth

So, g ∝ \(\frac{1}{R^2}\)               ...........3

From the equations 2 and 3, we get,

W ∝ \(\frac{1}{R^2}\)

Since R is minimum at poles, so weight is maximum at poles and similarly since R is maximum at equator, so weight is minimum at equator.

Hence, we weigh more at poles than at equator.

No, acceleration due to gravity do not depend on the mass of the body.

Explanation: The acceleration produced in a freely falling body by the gravitational pull of the earth is called the acceleration due to gravitation.

We know that,

Force = Mass × Acceleration

F = m × a

a = \(\frac{F}{m}\)         ........1

Where F is the force on the object of mass m dropped from a distance r from the centre of earth of mass M.

So, force exerted by the earth on the object is

F = G\(\frac{M*m}{r^2}\)          ...........2

M = Mass of Earth

m = mass of object

r = distance of object from centre of earth

Now, from equation 1 and 2,

a = G\(\frac{M*m}{r^2*m}\)

a = G\(\frac{M}{r^2}\)

Now, from above,

a = g = Acceleration due to gravity

We also see that, although force is depending on the mass of the object,

F = G\(\frac{M*m}{r^2}\)

But acceleration due to gravity is independent of the mass.

g = G\(\frac{M}{r^2}\)

We know that earth is not a perfect sphere. So, the value of R is not the same at all places.

So, the value of acceleration due to gravity g is not constant at all places on the surface of earth.

g = G\(\frac{M}{R^2}\)

Where, G = Gravitational Constant

M = Mass of Earth

R = Radius of Earth

Since, G and M are constant, so g depends on the radius of earth.

So, from the above we can say that,
g ∝ \(\frac{1}{R^2}\)

Now, radius of earth is minimum at poles. So, acceleration due to gravity is maximum at poles.

But radius at equator is maximum, so acceleration due to gravity is minimum at equator.

At the equator

Where the radius is small. Force of attraction is maximum at the poles.

No radius of the earth is not same everywhere.

If a body is being continuously raised from the surface of the earth, force of attraction decreases.

When it reaches the centre attractive force becomes zero.

The acceleration produced on a body falling freely under gravity is known as acceleration due to gravity.

SI unit – m s^-2 , Dimensional formula LT^-2.

When an object is held in the hand, the gravitational force acting on the object due to the earth is balanced by the person holding the object. When the object is released from the hand, it falls on the earth due to the earth’s gravitational force.

The value of g will be high at Goa Beach.

The minimum energy required by a satellite to escape from Earth ‘s gravitational influence is the binding energy of the satellite.

Some satellites that revolve around the Earth in equatorial plane have same sense of rotation as that of the Earth. The also have the same period of rotation as that of the Earth i.e.. 24 hours. Due to this, these satellites appear stationary from the Earth’s surface and are known as geostationary satellites.

The minimum velocity with which a both’ should he thrown vertically upwards from the surface of the Earth so that it escapes the Earth ‘s gravitational field, is called the escape velocity (v_e) of the body.

The gravitational force between the earth and the moon, the electromagnetic force between two charged particles in motion, the nuclear force between a proton and a neutron in the nucleus of an atom.

The weight of a body

(i) On the earth. W₁ = mg₁

(ii) on the moon, W₂ = mg₂

(iii) on Mars, W₃ = mg₃

∴ W₂ = W₁ × \(\frac{g_2}{g_1}\) and W₃ = W₁ × \(\frac{g_3}{g_1}\)

Now, g₁ = 9.81 m/s₂, g₂ = 1.67 m/s² and g₃ = 3.72 m/s²

If W₁ = 500 N,

W₂ = 500 × \(\frac{1.67}{9.81}\) N = 85.12N(approx.)

and W₃ = 500 × \(\frac{3.72}{9.81}\) N = 189.6 N (approx.)

All of these

EXPLANATION: 

Consider a thin spherical shell and choose r = infinity where potential is zero. At infinity V = 0, E = 0. So (a) is true. At any point outside the shell V ≠ 0 and E ≠ 0. So (d) is true. At any point inside the shell E = 0 but V ≠ 0. So (c) is true.

Keeping infinity as a reference point (b) cannot be true. But if we choose reference point just outside the shell then here V = 0 but E ≠ 0. So (b) is true.

B) Towards centre

Correct option is D) i, ii, iii

Gravitational force is the weakest force in nature. There are many properties of gravitational force, but first of all let us discuss about the various fundamental forces present in nature.

Following are the main features of gravitational force:

(i) It is always an attractive force.

(ii) It is independent of the medium between the
particles.

(iii) It holds good over a wide range of distances (i.e.,from interplanetary distances to interatomic distances).

(iv) It is an action-reaction pair i.e., the force of attraction exerted by body A on body B is equal to the force of attraction exerted by body B on body A. However, the acceleration of the two bodies will not be equal.

(v) The gravitational force between two particles
is independent of presence or absence of other particles.

(vi) The total gravitational force on one particle due to number of particles is the resultant of forces of attraction exerted on the given particle due to individual particles, i.e.,
It proves that the principle of superposition is valid.

(vii) It expresses, the force between point masses.

(viii) It is a conservative force, i.e., the work done in moving a particle once around a closed path under the action of gravitational force is zero

A ball is thrown upward. So,

Initial Speed = u

Distance = s = 100 m

Final Speed = v = 0

Acceleration = a = g = -9.8 ms^-2 (negative, since acceleration is opposite the motion)

Now,

v² = u² + 2as

0 = u² - 2 x 9.8  x 100

u = \(\sqrt {2\times9.8\times100}\)

u = 44.27 ms^-1

Hence, option B is correct.