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(b) is greatest when it is closest to the Earth

(b) increase by 0.5%

v ∝ \(\frac{1}{\sqrt{r}}\) , % increase in speed = \(\frac{1}{2}\)%decrease in radius = \(\frac{1}{2}\)(1%) = 0.5%

(b) Earth’s rotation 

Because Earth rotation from west to east direction.

Law of Conservation of angular momentum.

(d) 29

Time period does not depends upon the mass of satellite.

(i) It is consequence of law of conservation of angular momentum. 

(ii) The square of the time period of any planet about the Sun is proportional to the cube of the semi-major axis of the elliptical orbit.

Kepler’s third law of planetary motion led to the establishment of the Newton’s gravitational force between the two bodies which is the inverse-square rule.

(a) Kepler’s I Law (Law of Orbits) : Each planet revolves around the Sun in an elliptical orbit. The sun is situated at one foci of the ellipse. 

(b) Kepler’s II Law (Law of Areas) : The position vector of the planet from the Sun sweeps out equal area in equal interval of time. That is the areal velocity of the planet around the Sun is constant. 

(c) Kepler’s III Law (Law of Periods) : The square of the time period of any planet about the Sun is proportional to the cube of the semi-major axis of the elliptical orbit. 

(i) Mass 

(ii) Angular Momentum 

(iii) Total Energy

I Law : 

Every planet moves in an elliptical orbit around the sun with sun at one of the focus.

II Law: 

The line joining the planet to the sun sweeps out equal areas in equal interval of time.

III Law: 

The square of the period of any planet about the sun is proportional to the cube of the semi-major axis of the ellipse.

The acceleration due to gravity at a height ‘h’ from the surface of earth is given by, g₁ = g\([1-\frac {2h}{R}]\) where g is the acceleration due to gravity on the surface of earth and R is the radius of the earth.

Variation in acceleration due to gravity due to altitude is given by, g_h = g\(\left(\frac{R}{R + h}\right)^2\)

where,

g_h = acceleration due to gravity of an object placed at h altitude 

g = acceleration due to gravity on surface of the Earth 

R = radius of the Earth 

h = attitude height of the object from the surface of the Earth. Hence, acceleration due to gravity decreases with increase in altitude.

The acceleration due to gravity at a height ‘h’ from the surface of earth is given by, g₁= g[1-2h/R] where g is the acceleration due to gravity on the surface of earth and R is the radius of the earth.

Mass of the earth (M) is far greater than that of stone (m). So the acceleration acquired by earth is lesser and that by the stone is greater.

They reaches the ground at the same time. 

Reason: Acceleration due to gravity does not depend on the mass of the body. It will be the same for all

bodies falling to the earth, g = GM/R²

It is a universal constant and is not affected by any factors.

The value of the acceleration due to gravity, g, changes from place to place on the earth. It also varies with the altitude and depth below the earth’s surface. The factors affecting the’ value of g are the shape of the earth, altitude and depth below the earth’s surface.

(1) The earth is not perfectly spherical. It is somewhat flat at the poles and bulging at the equator. At the surface of the earth, the value of g is maximum (9.832 m/s²) at the poles as the polar radius is minimum, while it is minimum (9.78 m/s²) at the equator as the equatorial radius is maximum.

(2) As the height (h) above the earth’s surface increases, the value of g decreases. It varies as \(\frac{1}{(R+h)^2}\), where R is radius of the earth.

(3) In the interior of the earth, on average, the value of g is less than that at the earth’s surface. As the depth below the earth’s surface increases, the value of g decreases and finally it becomes zero at the centre of the earth.

At the equator

• Mass of the earth
• Radius of the earth

At the poles

Value of g be different everywhere on earth

No, the value of g is different at different places on the surface of earth. The shape of the earth is not exactly spherical, it is flattened at the poles and is bulging out at the equator. Due to which the radius of the earth is smaller at poles and is larger at equator. Since \(g\propto\frac{1}{R^2}\) , therefore the acceleration due to gravity is smaller at equator than that at poles.

The two particles will move on a circular path if they always remain diametrically opposite to that the gravitational force on one particle due to other is directly along the radius.

Taking into consideration the circulation of the one particle, we have

Centripetal force = Gravitational force

\(\frac{mv^2}{r}=\frac{GMm}{(2r)^2}\)

or v = \(\sqrt{\frac{GM}{4r}}\)

It is so because the gravitational attraction of earth provides the necessary centripetal force to the moon for its orbital motion around the earth. Due to which the moon is revolving around the earth.

Gravitational potential at a point in a gravitational field of a body is defined as the amount of work done in bringing a body of unit mass from infinity to that point without acceleration. Its S.I. Unit is J-kg^-1 .

Satellite is a smaller body revolving around another heavier body in stable orbit, e.g., moon is satellite of earth, earth is satellite of sun.

The velocity required by the body to escape from the gravitational pull of the earth is called the escape velocity.

v_e = \(\sqrt{2qr}=\sqrt{2}v_o\)

where v_o is the orbital velocity.

Aryabhatta, Bhaskara I, Rohini, APPLE, INSAT etc.

As escape velocity = \(\sqrt{\frac{2GM}{R}}\)

Therefore its values are different for different planets which are of different masses and different sizes.

It should be greater in night. In the day time the body is pulled by earth and sun in two opposite directions. This will result into decrease in weight. 

During night time the earth and the sun pull the body in the same direction so the weight will increase in night.

Molecules of hydrogen escape the earth factor because their R.M.S. velocity is more than that of oxygen molecules, R.M.S. velocity is given by

r_rms \(\propto\) \(\frac{1}{\sqrt m}\)

where m is the mass of the molecules.

Mass of oxygen molecules is larger than hydrogen so its R.M.S> velocity is less than that of hydrogen. Thus, it escapes slower than hydrogen.

The value of g is larger at the poles than at the equator. If the sugar is weighed in a physical balance then there will be no difference. If it is weighed by a spring balance, calibrated at the equator, then 1 kg of sugar will have a lesser amount at poles.

When a satellite stops orbiting, its kinetic energy becomes zero. This energy is converted into potential energy of the satellite. It then, starts falling towards the earth while crossing the atmosphere with great speed it may even burn.