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A comet with small velocity and high mass, doesn’t trigger the moon much. It just makes a circular shaped impact. The moon is ment for the protection for life on earth and to attain stability for the earth rotation. But a comet with large mass and with large velocity may destroy the moon completely or its impact makes the moon, go out of its orbit

(d) It is always zero 

2 & 3. No work is done on the earth revolving around it in perfectly circular orbit. The earth revolves around the sun due to gravitational force of attraction between the sun and the earth. This force around the sun. This centripetal force is always perpendicular to the linear displacement. 

Work done = W = F.d cos θ Since θ = 90°

= F.δ (0)  cos 90° = 0

W = 0

Correct answer is (a) Zero

Since there is a net force acting on the planet, its velocity changes which means its linear momentum changes. In fact, the absolute value of linear momentum changes too since the planet’s speed is variable as it goes around in its elliptical orbit . So the linear momentum of planet is not conserved. But the angular momentum about the sun is conserved. Since the torque of gravitational force is zero.

The total mechanical energy remains constant for an isolated system objects that interact with conservative forces. So, the total energy of the system of planet is conserved. The single energy of the planet is not conserved.

Correct answer is (b) 48 √2

Hint : T∝ \(r^\frac{3}{2}\) if r becomes double then time period will become (2)^{\(\frac{3}{2}\)} so the new time period will be 24 x 2√2 hours. i.e., 48√2 hours.

• Polar satellites are uses in weather and environment monitoring. 
• They are used in spying work for military purposes. 
• They are used to study topography of Moon, Venus and Mars

• In communicating radio, T.V and telephone signals across the world. 
• In studying upper regions of the atmosphere. 
• In forecasting weather. 
• In deter ming the exact shape and dimensions of the earth. 
• In studying solar radiations and cosmic rays.

Orbital velocity is the velocity required to put the satellite into its orbit around the earth.

Weight of a body is defined as the gravitational force with which a body is attracted towards the centre of the earth. Hence the weight of a body is given by w = mg (or ) \(\vec W = m \vec g\)

The Earth’s spinning motion can be proved by observing star’s position over a night. Due to Earth’s spinning motion, the stars in sky appear to move m circular motion about the pole star.

If the orbits of the Moon and Earth lie on the same plane, during full Moon of every month, we can observe lunar eclipse. If this is so dining new Moon we can observe solar eclipse. But Moon’s orbit is tilted 5° with respect to Earth’s orbit. Due to this 5° tilt, only during certain periods of the year, the Sun, Earth and Moon align in straight line leading to either lunar eclipse or solar eclipse depending on the alignment.

In a given datas tells us the relation connecting to the semi major axis is proportional to the two times of square of the time period. a ∝ 2T²

The force of attraction applied by earth is called its gravity.

The earth attracts (or pulls) all the objects towards its centre. The force with which it pulls objects towards itself is called gravitational force of earth or gravity.

Example:
A stone dropped from a height falls towards earth because the earth exerts a force of attraction (called gravity) on the stone and pulls it down.

According to Newton, every object in this universe attracts every other object with a certain force. The force with which two objects attract each other is called gravitation.

The gravitation acts between two objects even if the two objects are not connected by any means.

If the mass of the objects (or bodies) are small, then the gravitational force between them is also very small (which cannot be detected easily).

When an object is dropped from some height, a uniform acceleration is produced in it by the gravitational pull of earth and this acceleration does not depend upon the mass of falling body. This uniform acceleration is called acceleration due to gravity.

The acceleration due to gravity is represented by ‘g’ and its value is 9.8 ms^-2.

The value of g changes slightly from place to place.

g = \(\frac{GM}{R^2}\)

where,G = Gravitational constant

M = Mass of Earth

R = Radius of Earth

The earth attracts every object towards its centre with a certain force which depends on the mass of the body and the acceleration due to gravity at that place

The weight of the body is the force with which it is attracted towards the centre of the earth.

Force = Mass × Acceleration

Where, Force = Weight = W

Acceleration = Acceleration due to gravity = g

And, Mass = m

Thus, W = m × g

Also, weight is a vector quantity that is it has both magnitude as well as direction.

Pressure depends on two factors :

(i) Force applied 

(ii) Area of surface over which force acts

The weight of the body on the moon will be one-sixth that of the earth. Therefore, the weight of the body on the surface of the moon will be 1N.

The weight of the body is the force with which it is attracted towards the centre of the earth.

W = m × g

Weight of the body changes from place to place, as the value of g changes from place to place.

Acceleration due to gravity g is zero at the centre of earth, as at the centre of earth we are surrounded by equal masses in all the direction, hence equivalent gravitational force acting on us due to earth is zero and thus making acceleration due to gravity to zero.

So, W = m × g = 0

Hence, weight at the centre of earth is zero.

a) True 

b) False 

c) False 

d) False 

e) False

1. In the experiment performed to find the magnitude of gravitational constant (G), Cavendish balance is used.

2. The large spheres in the balance attract the nearby smaller spheres by equal and opposite force \(\overrightarrow{F}\) . 

Hence, a torque is generated without exerting any net force on the bar.

3. Due to the torque the bar turns and the suspension wire gets twisted till the restoring torque due to the elastic property of the wire becomes equal to the gravitational torque.

4. If r is the initial distance of separation between the centres of the large and the neighbouring small sphere, then the magnitude of the force between them is, F = \(G\frac{mM}{r^2}\)

5. If length of the rod is L, then the magnitude of the torque arising out of these forces is τ = FL = \(G\frac{mM}{r^2}\)L

6. At equilibrium, it is equal and opposite to the restoring torque.

∴ \(G\frac{mM}{r^2}\)L = Kθ

Where, K is the restoring torque per unit angle and θ is the angle of twist.

7. By knowing the values of torque τ₁ it and corresponding angle of twist a, the restoring torque per unit twist can be determined as K = τ₁/α.

8. Thus, in actual experiment measuring θ and knowing values of τ, m, M and r, the value of G can be calculated from equation (2).

Let,

u = Initial velocity of stone

H = Maximum height attained by the stone

v = final velocity at maximum height = 0

t = time taken to attain maximum height = 3 s

a = -g = acceleration due to gravity

(i) We know that,

v = u + at

0 = u – g × 3

u = 3g = 3 × 9.8 ms-2

u = 29.4 ms-1

(ii) From the equations of motion, we know that,

S = ut + \(\frac{1}{2}\)at²

H = 29.4 x 3 + \(\frac{1}{2}\)9.8 x 9

H = 88.2 + 44.1

H = 132.3 m

Data : m = 20 kg, g = 9.8 m/s²,

R = 6400 km = 6.4 × 10⁶ m

(i) The gravitational potential energy of the body = \(-\frac{GMm}{R}=-mgR\)

\((\because\,g=\frac{GM}{R^2})\)

= – 20 kg × 9.8 m/s² × 6.4 × 10⁶ m

= – 1.2544 × 10⁹ J.

(ii) To make the body free from the gravitational influence of the earth, it should be provided kinetic energy equal to 1.2544 × 10⁹ J.

Data : m = 20 kg, u = 100 m/s .

(i) The kinetic energy of the body = \(\frac{1}{2}\) mv² = \(\frac{1}{2}\) × 20 kg × (100 m/s)² = 10⁵ J.

(ii) The total energy of the body = kinetic energy + potential energy = 10⁵ J + (- 1.2544 × 10⁹ J)

= (1 – 12544) × 10⁵ J = – 12543 × 10⁵ J

= – 1.2543 × 10⁹ J.

Following is the difference between mass and weight of an object: 

a) The mass of an object is defined as the quantity of matter that is contained in it while weight of an object is the force with which the object gets attracted towards the center of the earth. 

b) The SI unit of mass is kg while the SI unit of weight is N. 

c) The mass of an object remains unchanged while weight of an object varies from place to place as it is dependent on the acceleration due to gravity.

Initial velocity, u = ? 

Final velocity, v = 0

Acceleration due to gravity, g = -10 m/s² 

Height, h = 5m 

a) For a freely falling body: 

v² = u² + 2gh 

u² = 100 

u = 10 m/s 

b) Using relation, v = u + gt 

t = 1s

Initial velocity, u=?

Final velocity, v=0

Acceleration due to gravity, g=-10m/s²

Height, h=5 m

(a) For a freely falling body:

v² = u² + 2gh

(0)² = u² + 2 x(-10)x 5

0= u² -100

u² = 100

So, u=10m/s

(b) Using relation, v=u + gt

0 = 10 + (-10) t

-10= -10 t

t=1sec

1. In terms of potential, we can write the potential energy of the Earth-mass system as, Gravitational potential energy (U) = Gravitational potential (V_r) × mass (m)

2. Thus, gravitational potential is gravitational potential energy per unit mass.

∴ V_r = \(\frac{U}{m}\)

3. For any conservative force field, the concept of potential can be defined on similar lines.

4. Gravitational potential difference between any two points in gravitational field can be written as,

V₂ – V₁ = \(\frac{U_2-U_1}{m}\)

= \(\frac{dW}{m}\)

This is the work done (or change in potential energy) per unit mass.

5. Therefore, in general, for a system of any two masses m₁ and m₂, separated by distance r, we can write,

U = \(-\frac{Gm_1\,m_2}{r}\)

= (V₁)m₂

= (V₂)m₁

Here V₁ and V₂ are gravitational potentials at r due to m₁ and m₂ respectively.

Given: m = 80 kg, U = 5 × 10⁹ J

To find: Gravitational potential (V)

Formula: V = \(\frac{U}{m}\)

Calculation: From formula,

V = \(\frac{5\times10^9}{80}=\frac{25}{4}\) × 10⁷

= 6.25 × 10⁷ J kg^-1

Potential of the body at the surface of the Earth is 6.25 × 10⁷ J kg^-1.

An object which revolves in an orbit around a planet is called as satellite.

Example:

• Moon is a natural satellite of the Earth.
• INSAT is an artificial satellite of the Earth.

1. Polar Satellites are placed in lower polar orbits.

2. They are at low altitude 500 km to 800 km.

3. Period of revolution of polar satellite is nearly 85 minutes, so it can orbit the Earth 16 time per day.

4. They go around the poles of the Earth in a north-south direction while the Earth rotates in an east-west direction about its own axis.

5. The polar satellites have cameras fixed on them. The camera can view small stipes of the Earth in one orbit. In entire day the whole Earth can be viewed strip by strip.

6. Polar region and equatorial regions close to it can be viewed by these satellites.

7. Polar satellites are used for weather forecasting and meteorological purpose. They are also used for astronomical observations and study of Solar radiations.

(1) Acceleration due to gravity at the surface of planet 

(2) Surface temperature of the planet