Samacheer Class 11 Physics - Gravitation

• Law of Orbits: Each planet revolves moves around the Sun in an elliptical orbit with the Sun at one of the foci of the ellipse.
• Law of area: The radial vector line joining the Sun to a planet sweeps equal areas in equal intervals of time.
• Law of period: The square of the time period of revolution of a planet around the Sun in its elliptical orbit is directly proportional to the cube of the semi-major axis of the ellipse.

The torque experienced by the Earth due to the gravitational force of the Sun is given by,
It implies that angular momentum \(\vec { L }\) is constant vector. Hence L is conserved.

Potential energy of a body at a point in a gravitational field is the work done by an external agent in moving the body from infinity to that point.

No, potential energy is not the property of a single system. It is due to the position of the object. Because, potential energy depends on two masses and the distance between them.

The escape speed is independent of the direction in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the same initial speed to escape Earth’s gravity force. This can be written as, v e = \(\sqrt{2 g R_{E}}\).

The negative sign of the total energy implies that satellite is bound to the Earth’s gravitational force. So it cannot escape from the Earth. At large distances satellite is not bound to Earth. It is completely free from gravitational force of the Earth.

Geostationary Satellite: It is the satellite which appears at a fixed position and at a definite height to an observer on earth.
Polar Satellite: It is the satellite which revolves in polar orbit around the earth.

The weight is defined as the downward force whose magnitude ‘w ’ is equal to that of upward force that must be applied to the object to hold it at rest or at constant velocity relative to the Earth.

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.

When the position of a star is observed over a night, it appears to move in circular motion about the pole star. This fact proves that Earth itself is spinning.
III. Long Answer Questions:

(i) Because of inverse dependence on square of the distance the distance between two masses increases, the strength of the force tends to decrease.
(ii) Because Uranus is at larger distance from the Sun compared to the Earth, the planet Uranus experiences less gravitational force from the Sun than the Earth.
(iii) The gravitational forces between two particles always constitute an action- reaction pair. From this, it is known that gravitational force exerted by the Sun on the Earth is always towards the Sun. The reaction-force is exerted by the Earth on the Sun. The direction of reaction is towards Earth.
(iv) The torque experienced by the Earth due to the gravitational force of the Sun is given by,
It implies that angular momentum \(\vec { L }\) is a constant vector. The angular momentum of Earth about the Sun is constant throughout the motion. It is true for all the planets.
(v) The expression \(\vec { F }\) = \(\frac{\mathrm{GM}_{1} \mathrm{M}_{2}}{r^{2}} \hat{r}\). Here it was assumed that both Earth and Sun are treated as point masses.
(vi) This assumption is a good approximation because the distance between the two bodies is very much larger than their diameters.
(vii) The force of attraction between a hollow sphere of mass M with uniform density and point mass m kept outside the hollow sphere is calculated by replacing the hollow sphere of mass M as equivalent to a point mass M located at the centre of the hollow sphere.
The force of attraction between the hollow sphere of mass M and point mass ‘m’ can be calculated by considering the hollow sphere also as another point mass. It is shown in the Figure below.
(viii) If we place another object of mass ‘m ’ inside a hollow sphere of mass M as in Figure below, the force experienced by this mass ‘m ’ will be zero.
(ix) The success of the law of gravitation is that it concludes that the mango that is falling down and the Moon orbiting the Earth are due to the same gravitational force.

Newton considered the orbits of the planets as circular. For circular orbit of radius r, the centripetal acceleration towards the centre is
Here v is the velocity and r, the distance of the planet from the centre of the orbit.
The velocity in terms of known quantities r and T, is
v = \(\frac { 2πr }{ T }\) … (2)
Here T is the time period of revolution of the planet. Substituting this value of ‘v’ in equation (1) we get,
Substituting the value of ‘a’ from (3) in Newton’s second law, F = ma where ‘m’ is the mass of the planet.
F = – \(\frac{4 \pi^{2} m r}{\mathrm{r}^{2}}\) … (4)
From Kepler’s third law,
\(\frac{r^{3}}{\mathrm{~T}^{2}}\) = k (constant) … (5)
\(\frac{r}{\mathrm{~T}^{2}}\) = \(\frac{k}{r^{2}}\) … (6)
By using equation (6) in the force expression, we can arrive at the law of gravitation.
F = – \(\frac{4 \pi^{2} m k}{\mathrm{r}^{2}}\) … (7)
Here negative sign implies that the force is attractive and it acts towards the centre. He equated the constant 4π²k to GM which turned out to be the law of gravitation.
F = – \(\frac{\mathrm{GMm}}{r^{2}}\)

The gravitational force experienced by the apple due to Earth.
F = – \(\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{M}_{\mathrm{A}}}{\mathrm{R}^{2}}\)
M A – Mass of the apple
M E – Mass of the Earth
R – Radius of the Earth.
Equating the above equation with Newton’s second law, we get
M A a A = – \(\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{M}_{\mathrm{A}}}{\mathrm{R}^{2}}\)
Simplifying the above equation we get,
a A = – \(\frac{\mathrm{GM}_{\mathrm{E}}}{\mathrm{R}^{2}}\)
a A is the acceleration of apple that is equal to ‘g’.
Similarly the force experienced by Moon due to Earth is given by
F = – \(\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{Mm}}{\mathrm{R}_{m}^{2}}\)
R m – distance of the Moon from the Earth.
M m – Mass of the Moon.
The acceleration experienced by the Moon is given by,
a m = – \(\frac{\mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{m}^{2}}\)
The ratio between the apple’s acceleration to Moon’s acceleration is given by
\(\frac{a_{\mathrm{A}}}{a_{m}}=\frac{\mathrm{R}_{m}^{2}}{\mathrm{R}^{2}}\)
From the Hipparchrus measurement, the distance to the Moon is 60 times that of Earth radius.
R m = 60R.
\(\frac{a_{\mathrm{A}}}{a_{m}}=\frac{(60 \mathrm{R})^{2}}{\mathrm{R}^{2}}\)
The apple’s acceleration is 3600 times the acceleration of the Moon.
The same result was obtained by Newton using his gravitational formula.

Consider two masses m 1 and m 2 are initially separated by a distance r’. m 1 is assumed to be fixed in its position.
To move the mass m 2 through an infinitesimal displacement \(\vec { dr }\) from \(\vec { r }\) to \(\vec { r }\) + \(\vec { dr }\) (shown in the Figure above), work has to be done externally. This infinitesimal work is given by
Hence the total work done for displacing the particle from r’ to r is

Let us consider the Earth and mass system, with r, the distance between the mass ‘nr and the Earth’s centre. Then the gravitational potential energy,
U = \(\frac{\mathrm{GM}_{e} m}{r}\) … (1)
Here r = R e + h, where R e is the radius of the Earth, h is the height above the Earth’s surface.
U = – G \(\frac{\mathrm{M}_{e} m}{\left(\mathrm{R}_{e}+h\right)}\) … (2)
If h << R e, equation (2) can be modified as
By using Binomial expansion and neglecting the higher order terms, we get
U = – \(\mathrm{G} \frac{\mathrm{M}_{e} m}{\mathrm{R}_{e}}\left(1-\frac{h}{\mathrm{R}_{e}}\right)\) … (4)
We know that, for a mass m on the Earth’s surface,
G\(\frac{\mathrm{M}_{e} m}{\mathrm{R}_{e}}\) = mgR e … (5)
Substituting equation (5) in (4) we get,
U = mgR e + mgh … (6)
In the equation (6) the first term can be omitted or taken to zero. Thus it can be stated that the gravitational potential energy stored in the particle of mass m at a height h from the surface of the Earth is U = mgh.

Freely falling bodies experience gravitational force only. As they fall freely, they are not in contact with any surface. When a lift falls with downward acceleration a = g, the person inside the lift is in the state of free fall or weightlessness. The normal force acting on the object is zero. In this case, the downward acceleration is equal to the acceleration due to the gravity of the Earth. i.e., (a = g). From equation,
N = m(g – a)
we get,
a= g
∴ N = m(g – g) = 0.
This is called the state of weightlessness.
Example: When the lift falls (when the lift wire cuts) with downward acceleration a = g, the person inside the elevator is in the state of weightlessness or free fall.

Consider an object of mass M on the surface of the Earth. When it is thrown up with an initial speed v i the initial total energy of the object is
E i = \(\frac { 1 }{ 2 }\)Mv i ² – \(\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) … (1)
Where, M E is the mass of the Earth and R E – the radius of the Earth. The term \(\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\) is the potential energy of the mass M.
When the object reaches a height far away from Earth and as approaching infinity, the gravitational potential energy becomes zero [U(∞) = 0] and the kinetic energy becomes zero as well. Hence the final total energy of the object becomes zero. It holds good for minimum energy and for minimum speed to escape. Otherwise Kinetic energy can be non-zero.
E f = 0
According to the law of energy conservation,
E i = E f … (2)
Substituting (1) in (2) we get,
The minimum speed required by an object to escape Earth’s gravitational field is escape speed. Hence v i is replaced v e. i.e.,

An object is on the surface of the Earth, experiences a centrifugal force that depends on the latitude of the object on Earth. If the Earth were not spinning, the force on the object would have been mg. An additional centrifugal force due to spinning of the Earth, is experienced by the object.
This centrifugal force is given by mω²R’.
R’= R cos λ,
where λ is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to acceleration due to gravity is

Let us consider an object of mass m at a height h from the surface of the Earth. Acceleration experienced by the object due to Earth is
If h << Re
Using Binomial expansion and taking the terms upto first order.
We find that g’

Let us consider a particle of mass m which is in a deep mine on the Earth. Let the depth of the mine as d. To calculate g’ at a depth d, consider the following points.
The part of the Earth that is above the radius (R e – d) do not contribute to the acceleration. The result is proved earlier and is given as GM
g’ = \(\frac{\mathrm{GM}^{\prime}}{\left(\mathrm{R}_{e}-d\right)^{2}}\)
M’ is the mass of the Earth of radius (R e – d)
Assuming the density of Earth p to be constant,
ρ = \(\frac { M }{ V }\)
Where M is the mass of the Earth and V its volume, Thus,
Here also g'< g. As depth increases, g’ decreases.

The distance covered by the satellite during one rotation in its orbit is equal to 2π(R E + h) and time taken for it is the time period, T. Then
Equation (3) implies that a satellite orbiting the Earth has the same relation between time and distance as that of Kepler’s law of planetary motion. For a satellite orbiting near the surface of the Earth, h is negligible compared to the radius of the Earth R E. Then,

The total energy of the satellite is the sum of its kinetic energy and the gravitational potential energy. The potential energy of the satellite is,
U = – \(\frac{\mathrm{GM}_{\mathrm{S}} \mathrm{M}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}\) … (1)
Here M s – Mass of the satellite, M E – Mass of the Earth, R E – Radius of the Earth.
The Kinetic energy of the satellite is
K.E = \(\frac { 1 }{ 2 }\)M s … (2)
Here v is the orbital speed of the satellite and is equal to
v = \(\sqrt{\frac{\mathrm{GM}_{\mathrm{E}}}{\left(\mathrm{R}_{\mathrm{E}}+h\right)}}\) … (3)
Substituting the value of v in (2), the kinetic energy of the satellite becomes,

Geo-stationary satellite is the satellite which revolves around the Earth with same speed and in the same direction as the Earth rotates.
R E + h = \(\left(\frac{\mathrm{GM}_{\mathrm{E}} \mathrm{T}^{2}}{4 \pi^{2}}\right)\) … (1)
By substituting the values for the time period, mass and radius of the Earth, in equation (1) h turns out to be 36,000 km. Such satellites are called “geo – stationary satellites”, since they appear to be stationary when seen from Earth. INSAT group of satellites that are basically geo – stationary satellites used for the purpose of telecommunication.
A Polar satellite covers a small strip of area from pole to pole during one revolution. In the next revolution it covers a different strip of area. Since the Earth would have moved by a small angle. In this way polar satellites cover the entire surface area of the Earth.

(i) During the observation of motion of the planets it is in the night sky by naked eyes over a period of a few months, it can be seen that the planets move eastwards and reverse their motion for a while and return to eastward motion again. This is called “retrograde motion” of planets.
(ii) Ptolemy introduced the concept of “epicycle” in his geocentric model. To explain this retrograde motion, according to this theory, while the planet orbited the Earth, it also underwent another circular motion termed as “epicycle”. A combination of epicycle and circular motion around the Earth gave rise to retrograde motion of the planets with respect to Earth.
(iii) But Ptolemy’s model became more and more complex as every planet was found to undergo retrograde motion.
(iv) According to heliocentric model, the Sun is at the centre of the solar system and all planets orbited the Sun. The retrograde motion of planets ‘with respect to Earth is due to the relative motion of the planet with respect to Earth. The retrograde motion from the heliocentric point of view.
The Earth orbits around the Sun faster than Mars. Due to the relative motion between Mars and Earth, Mars appears to move backwards from July to October. In the same way the retrograde motion of all other planets was explained successfully by the Copemicus model. The heliocentric model slowly replaced the geocentric model, because of its simplicity.

Eratosthenes observed that during noon time of summer solstice the Sun’s rays cast no shadow in the city Syne. It was located 500 miles away from Alexandria. At the same day and same time he found that in Alexandria the Sun’s rays made an angle 7.2 degree with local vertical as shown in the figure. He realized that this difference of 7.2 degree was due to the curvature of the Earth.
The angle 7.2 degree is equivalent to \(\frac { 1 }{ 8 }\) radian. So θ = \(\frac { 1 }{ 8 }\) rad.
If S is the length of the arc between the cities of Syne and Alexandria, and if R is radius of Earth, then,
S = Rθ = 500 miles
So, the radius of the Earth.
R = \(\frac { 500 }{ θ }\) miles
R = 500 \(\frac { miles }{ 1/8 }\)
R = 4000 miles
1 mile is equal to 1.609 km. So, he measured the radius of the Earth to be equal to R = 6436km, which is amazingly close to the correct value of 6378km.

It is possible to measure the radius of the shadow of the Earth at the point where the Moon crosses. When the Moon exits from the umbra shadow, it appears red in color. As soon as the Moon exits from the umbra shadow, it appears in a crescent shape.
By knowing the apparent radii of the Earth’s umbra shadow and the Moon, the ratio of these radii can be calculated.
The apparent radius of Earth’s umbra shadow = R s = 13.2cm.
The apparent radius of the Moon = R m = 5.15 cm.
The ratio \(\frac{\mathrm{R}_{s}}{\mathrm{R}_{m}}\) ≈ 2.56
The radius of the Earth’s umbra shadow is,
R s = 2.56 x R m
The radius of Moon R m = 1737km
The radius of the Earth’s umbra shadow is,
R s = 2.56 x 1737 km ≅ 4446km.
The correct radius is 4610km.
IV. Conceptual Questions:

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.

From the photograph, the apparent radii of the Earth’s umbra shadow and the radius of the Moon, and the apparent radius of Earth can be calculated. By looking at the shape of the shadow caused by Earth on the Moon during lunar eclipse. We can confirm that Earth is spherical in shape.
V. Numerical Problems:

Given: a 1 = 2a 1
To find: T 2 = ??
Formula: T² ∝ a³ => T ∝ (a) 3/2
T 2 ∝ (2a 1 ) 3/2
∝ (2) 3/2 (a 1 ) 3/2
T2 ∝ 2\(\sqrt{2}\) T 1
∴ T 2 = (2/\(\sqrt{2}\))T 1.

Time period T² ∝ a³
T 1 = 2; a 1 = 8
a 1 = 2² x 2 = 2T 1 (T 1 = 2)
a 2 = 3² x 2 = 2T 2 ² (T 2 = 3)
a 3 = 4² x 2 = 2T 3 ² (T 3 = 4)
a 4 = 5² x 2 = 2T 4 ² (T 4 = 5)
a 5 = 6² x 2 = 2T 5 ² (T 5 = 6)
∴ a ∝ 2T²

Incorrect value of G = 6.67 x 10 11
If mass m = 1
then weight W’ = mg’- 9.8 x 10 22
W’ = 9.8 x 10 22 W

Given: r = 150 million km
r = 150 x 10 6 km
= 150 x 10 9 m
M e = 5.9 x 10 24 kg
M s = 1.9 x 10 30 kg
To find: Potential energy = ??
Formula:

Given: V=30km/s
Formula:
\(\left.\begin{array}{l}
\text { Kinetic } \\
\text { energy(K.E) }
\end{array}\right\}\) = \(\frac { 1 }{ 2 }\)mv²
= \(\frac { 1 }{ 2 }\) x 5.9 x 10 24 x 30 x 30 x 10 32
= \(\frac { 1 }{ 2 }\) x 5.9 x 3 x 3 x 10 32
K.E = 26.55 x 10 32 J.
∴ T.E = K.E + P.E
T.E = 26.55 x 10 32 – 49.84 x 10 32
T.E = – 23.29 x 10 32 J.
Total energy is negative. It implies Earth is bounded with Sun.

g’ = g – ω² Rcos² λ
Latitude of g at Chennai λ = 13°
ω²R = ( \(\frac { 1 }{ 2 }\) )²R
= \(\left(\frac{2 \times 3.14}{24 \times 60 \times 60}\right)^{2}\) x 6400 x 10³
ω²R = 3.4 x 10 -2 ms -2
13°= 0.2268 rad
g Chennai = 9.8 – (3.4 x 10 -2 ) x (cos 0.2268) 2
g Chennai = 9.7667 ms -2 = 9.767 ms -2
Latitude of g at Kanyakumari = 8.08°