NEET Physics Class 11 Chapter 7 Gravitation Notes

Gravitation Introduction

The motion of celestial bodies such as the sun, the moon, the earth the planets etc. has been a subject of fascination since time immemorial.

• Indian astronomers of ancient times have done brilliant work in this field, the most notable among them being Arya Bhatt the first person to assert that all planets including the Earth revolve around the sun.
• A millennium later the Danish astronomer Tycobrahe (1546-1601) conducted a detailed study of planetary motion which was interpreted by his pupil Johnaase Kepler (1571-1630), ironically after the master himself had passed away.
• Kepler formulated his important findings in three laws of planetary motion

Universal Law Of Gravitation: Newton’s Law

According to this law, “Each particle attracts every other particle. The force of attraction between them is directly proportional to the product of their masses and inversely proportional to the square of the distance between them”.

⇒ \(\mathrm{F} \propto \frac{m_1 m_2}{r^2}\)

or \(\mathrm{F}=G \frac{m_1 m_2}{r^2}\)

where G = 6.67 × 10^-11 Nm² kg^-2 is the universal gravitational constant. This law holds good irrespective of the nature of two objects (size, shape, mass etc.) at all places and at all times. That is why it is known as a universal law of gravitation.

Dimensional Formula Of G:

G = \(\frac{F r^2}{m_1 m_2}\)

⇒ \(\frac{\left[M L T^{-2}\right]\left[L^2\right]}{\left[M^2\right]}\)

= [M^-1 L³ T^-2]

Newton’s Law Of Gravitation In Vector Form:

⇒ \(\overrightarrow{\mathrm{F}}_{12}=\frac{G m_1 m_2}{r^2} \hat{r}_{12}\)

⇒ \(\overrightarrow{\mathrm{F}}_{21}=\frac{G m_1 m_2}{r^2}\)

Where \(\overrightarrow{\mathrm{F}}_{12}\) is the force on mass m₁ exerted by mass m₂ and vice-versa.

⇒ \(\overrightarrow{\mathrm{F}}_{12}=\frac{G m_1 m_2}{r^2} \hat{r}_{12}\)

⇒ \(\overrightarrow{\mathrm{F}}_{21}=\frac{G m_1 m_2}{r^2} \hat{\mathrm{r}}_{21}\)

Now \(\hat{r}_{12}=-\hat{r}_{21}\)

Thus \(\vec{F}_{21}=\frac{-G m_1 m_2}{r^2} \hat{r}_{12}\) Comparing above, we get \(\vec{F}_{12}=-\vec{F}_{21}\)

Important Characteristics Of Gravitational Force

1. Gravitational force between two bodies forms an action and reaction pair i.e. the forces are equal in magnitude but opposite in direction.
2. Gravitational force is a central force i.e. it acts along the line joining the centres of the two interacting bodies.
3. Gravitational force between two bodies is independent of the nature of the medium, in which they lie.
4. The gravitational force between two bodies does not depend upon the presence of other bodies.
5. Gravitational force is negligible in the case of light bodies but becomes appreciable in the case of massive bodies like stars and planets.
6. Gravitational force is long range-force i.e., gravitational force between two bodies is effective even if their separation is very large. For example, the gravitational force between the sun and the earth is of the order of 10²⁷ N although the distance between them is 1.5 × 10⁷ km

Question 1. The centres of two identical spheres are at a distance of 1.0 m apart. If the gravitational force between them is 1.0 N, then find the mass of each sphere. (G = 6.67 × 10^-11 m³ kg^-1 sec^-1)
Answer:

Gravitational force F = \(\frac{G m \cdot m}{r^2}\)

on substituting F = 1.0 N , r = 1.0 m and G = 6.67 × 10^-11 m³ kg^-1 sec^-1

we get m = 1.225 × 10⁵ kg

Principle Of Superposition

The force exerted by a particle or other particle remains unaffected by the presence of other nearby particles in space.

The total force acting on a particle is the vector sum of all the forces acted upon by the individual masses when they are taken alone.

⇒ \(\vec{F}_1=\vec{F}_1+\vec{F}_2+\vec{F}_3+\ldots \ldots\)

Question 2.

Four point masses each of mass ‘m’ are placed on the corner of a square of side ‘a’. Calculate the magnitude of gravitational force experienced by each particle.

Answer:

F_r= resultant force on each particle = 2F cos 45º + F₁

⇒ \(\frac{2 G \cdot m^2}{a^2} \cdot \frac{1}{\sqrt{2}}+\frac{G m^2}{(\sqrt{2} a)^2}=\frac{G \cdot m^2}{2 a^2}(2 \sqrt{2}+1)\)

Gravitational Field

The space surrounding the body within which its gravitational force of attraction is experienced by other bodies is called the gravitational field.

• A gravitational field is very similar to an electric field in electrostatics where charge ‘q’ is replaced by mass ‘m’ and electric constant ‘K’ is replaced by gravitational constant ‘G’.
• The intensity of the gravitational field at a point is defined as the force experienced by a unit mass placed at that point.

⇒ \(\vec{E}=\frac{\vec{F}}{m}\)

The unit of the intensity of the gravitational field is N kg^-1.

Intensity of gravitational field due to point mass:

The force due to mass m on test mass m0 placed at point P is given by :

⇒ \(\mathrm{F}=\frac{\mathrm{GMm}_0}{\mathrm{r}^2}\)

Hence \(E=\frac{F}{m_0}\)

⇒ \(E=\frac{G m}{r^2}\)

In vector form \(\vec{E}=-\frac{G M}{r^2} \hat{r}\)

Dimensional formula of intensity of gravitational field = \(\frac{F}{m}=\frac{\left[M L T^{-2}\right]}{[M]}\)

⇒ \(\left[\begin{array}{ll}M^0 & L T^{-2}\end{array}\right]\)

Question 1. Find the relation between the gravitational field on the surface of two planets A and B of masses m_A, mB and radius R_A and R_B respectively if

1. They have equal mass
2. They have equal (uniform) density

Answer: Let EA and EB be the gravitational field intensities on the surface of planets A and B.

then, \(\mathrm{E}_{\mathrm{A}}=\frac{G m_A}{R_A^2}=\frac{G \frac{4}{3} \pi R_A^3 \rho_A}{R_A^2}\)

⇒ \(\frac{4 G \pi}{3} \rho_A R_A\)

Similarly, \(E_B=\frac{G m_B}{R_{B^2}}=\frac{4 G}{3} \pi \rho_{\mathrm{B}} \mathrm{R}_{\mathrm{B}}\)

For \(m_A=m_B\)

⇒ \(\frac{E_A}{E_B}=\frac{R_B^2}{R_A^2}\)

For \(\rho_{\mathrm{A}}=\rho_{\mathrm{B}}\)

⇒ \(\frac{E_A}{E_B}=\frac{R_A}{R_B}\)

Gravitational Potential

The gravitational potential at a point in the gravitational field of a body is defined as the amount of work done by an external agent in bringing a body of unit mass from infinity to that point, slowly (no change in kinetic energy). Gravitational potential is very similar to electric potential in electrostatics.

Gravitational Potential Due To A Point Mass :

Let the unit mass be displaced through a distance dr towards mass M, then work done is given by

dW = F dr = \(\frac{\mathrm{Gm}}{\mathrm{r}^2} \mathrm{dr}\)

The total work done in displacing the particle from infinity to point P is −

⇒ \(\mathrm{W}=\int d W=\int_{\infty}^r \frac{G M}{r^2} d r=\frac{-G M}{r}\)

Thus gravitational potential, \(V=-\frac{G M}{r}\)

The unit of gravitational potential is J kg^-1. Dimensional Formula of gravitational potential

⇒ \(\frac{\text { work }}{\text { mass }}=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{[\mathrm{M}]}\)

= [M°L² T^-2].

Question 1. Find out the potential at P and Q due to the two-point mass system. Find out work done by external agents in bringing unit mass from P to Q. Also find work done by gravitational force.

Answer: V_P1 = potential at P due to mass ‘m’ at ‘l’ = –\(\frac{G m}{\ell}\)

⇒ \(\mathrm{V}_{\mathrm{P} 2}=-\frac{G m}{\ell}\)

∴ \(\mathrm{V}_{\mathrm{p}}=\mathrm{V}_{\mathrm{P} 1}+\mathrm{V}_{\mathrm{P} 2}=-\frac{2 G m}{\ell}\)

⇒ \(V_{Q 1}=-\frac{G M}{\ell / 2}\)

⇒ \(V_{Q 2}=-\frac{G m}{\ell / 2}\)

∴ \(\mathrm{v}_{\mathrm{Q}}=\mathrm{v}_{\mathrm{Q} 1}+\mathrm{v}_{\mathrm{Q} 2}=-\frac{G m}{\ell / 2}-\frac{G m}{\ell / 2}=-\frac{4 G m}{\ell}\)

Force at point Q = 0

Work done by external agent = \(\left(\mathrm{V}_{\mathrm{Q}}-\mathrm{V}_{\mathrm{P}}\right) \times 1=-\frac{2 G M}{\ell}\)

Work done by gravitational force = V_P– V_Q = \(\frac{2 G M}{\ell}\)

Question 2. Find the potential at a point ‘P’ at a distance ‘x’ on the axis away from the centre of a uniform ring of mass M and radius R.
Answer:

The ring can be considered to be made of a large number of point masses (m₁, m₂ ………. etc) Ring

⇒ \(\mathrm{V}_{\mathrm{p}}=-\frac{G m_1}{\sqrt{R^2+x^2}}-\frac{G m_2}{\sqrt{R^2+x^2}}-\ldots \ldots .\)

⇒ \(-\frac{G}{\sqrt{R^2+x^2}}\left(m_1+m_2 \ldots . .\right)=-\frac{G M}{\sqrt{R^2+x^2}}\) where, M = m₁+m₂+m₃+……..

Potential at centre of ring = \(-\frac{G \cdot M}{R}\)

Relation Between Gravitational Field And Potential

The work done by an external agent to move unit mass from one point to another point in the direction of the field E, slowly through an infinitesimal distance dr = Force by external agent × distance moved = – Edr.

Thus dV = – Edr

⇒ E = – \(\frac{d V}{d r}\)

Therefore, the gravitational field at any point is equal to the negative gradient at that point.

Uniform Solid Sphere

Point P inside the shell. r ≤ a, then

V = \(-\frac{\mathrm{Gm}}{2 \mathrm{a}^3}\left(3 \mathrm{a}^2-\mathrm{r}^2\right)\)

and E = \(-\frac{G M r}{a^3}\) ,and at the centre V = \(-\frac{3 G M}{2 a}\) and E = 0

Point P outside the shell. r > a, then V = \(-\frac{G M}{r}\) and E = \(-\frac{G M}{r^2}\)

Uniform Thin Spherical Shell

Point P Inside the shell. r ≤ a , then V = \(-\frac{G M}{a}\) and E = 0

Point P outside shell. r ≥ a, then V = \(-\frac{G M}{r}\) and E = \(-\frac{G M}{r^2}\)

Gravitational Potential Energy

The gravitational potential energy of two mass systems is equal to the work done by an external agent in assembling them, while their initial separation was infinity.

Consider a body of mass m placed at a distance r from another body of mass M. The gravitational force of attraction between them is given by

⇒ \(\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{r}^2}\)

Now, Let the body of mass m is displaced from a point. C to B through a distance ‘dr’ towards the mass M, then work done by internal conservative force (gravitational) is given by,

⇒ \(\mathrm{dW}=\mathrm{F} \mathrm{dr}=\frac{G M m}{r^2} \mathrm{dr}\)

⇒ \(\int d W=\int_{\infty}^r \frac{G M m}{r^2} \mathrm{dr}\)

∴ Gravitational potential energy, \(\mathrm{U}=-\frac{\mathrm{GMm}}{r}\)

Increase In Gravitational Potential Energy:

Suppose a block of mass m on the surface of the earth. We want to lift this block by ‘h’ height. Work required in this process = increase in P.E. = U_f– U_i= m(V_f– V_i)

⇒ \(\mathrm{W}_{\mathrm{ext}}=\Delta \mathrm{U}=(\mathrm{m})\left[-\left(\frac{G M_c}{R_e+h}\right)-\left(-\frac{G M_c}{R_e}\right)\right]\)

W_ext = ΔU = \(\mathrm{GM}_e \mathrm{~m}\left(\frac{1}{R_e}-\frac{1}{R_e+h}\right)=\frac{G M_e m}{R_e}\left(1-\left(1+\frac{h}{R_e}\right)^{-1}\right)\)

(as h << R_e, we can apply the Binomial theorem)

W_ext = ΔU = \(\frac{G M_e m}{R_e}\left(1-\left(1-\frac{h}{R_e}\right)\right)=(\mathrm{m})\left(\frac{G M_e}{R_e^2}\right) \mathrm{h}\)

W_ext = ΔU = mgh

This formula is valid only when h << R_e

Question 1. A body of mass m is placed on the surface of the earth. Find the work required to lift this body by a height

1. h = \(\frac{R_c}{1000}\)
2. h = R_e

Answer: h = \(\frac{R_c}{1000}\) as h << R_e , so

we can apply

W_ext = \(\mathrm{W}_{\mathrm{ext}}=\mathrm{U} \uparrow=\mathrm{mgh}\)

⇒ \(\mathrm{W}_{\mathrm{ext}}=\left(\frac{G M_e}{R_e{ }^2}\right)\left(\frac{R_e}{1000}\right)(\mathrm{m})=\frac{G M_e m}{1000 R_e}\)

h = R_e, in this case, h is not much less than R_e, so we cannot apply ΔU = mgh so we cannot apply ΔU = mgh

W_ext = U↑ = U_f– U_i= m(V_f– V_i)

⇒ \(\mathrm{W}_{\mathrm{ext}}=\mathrm{m}\left[\left(-\frac{G M_e}{R_e+R_e}\right)-\left(-\frac{G M_e}{R_e}\right)\right]\)

⇒ \(\mathrm{W}_{\mathrm{ost}}=-\frac{G M_e m}{2 R_e}\)

Acceleration Due To Gravity

It is the acceleration, a freely falling body near the earth’s surface acquires due to the earth’s gravitational pull.

• The property by virtue of which a body experiences or exerts a gravitational pull on another body is called gravitational mass mG, and the property by virtue of which a body opposes any change in its state of rest or uniform motion is called its inertial mass
• m₁ thus if \(\overrightarrow{\mathrm{E}}\) is the gravitational field intensity due to the earth at a point P, and g is acceleration due to gravity at the same point, then \(m_1 \vec{g}=m_G \vec{E}\)
• Now the value of inertial and gravitational mass happens to be exactly the same to a great degree of accuracy for all bodies. Hence,\(\vec{g}=\vec{E}\)

The gravitational field intensity on the surface of the earth is therefore numerically equal to the acceleration due to gravity (g), there. Thus we get,

⇒ \(g=\frac{G M_e}{R_e^2}\)

where, M_e= Mass of earth

R_e = Radius of earth

Note: Here the distribution of mass in the earth is taken to be spherical and symmetrical so that its entire mass can be assumed to be concentrated at its centre for the purpose of calculation of g.

Variation Of Acceleration Due To Gravity

Effect of Altitude

Acceleration due to gravity on the surface of the earth is given by,

g = \(\frac{G M_e}{R_e^2}\)

Now, consider the body at a height ‘h’ above the surface of the earth, then the acceleration due to gravity at height ‘h’ given by

⇒ \(\mathrm{g}_{\mathrm{h}}=\frac{G M_e}{\left(R_e+h\right)^2}=\mathrm{g}\left(1+\frac{h}{R_e}\right)^{-2} \simeq \mathrm{g}\left(1-\frac{2 h}{R_e}\right)\)

when h << R.

The decrease in the value of ‘g’ with height h = g – g_h= \(\frac{2 g h}{R_e}\)

The percentage decrease in the value of \(‘ \mathrm{~g}^{\prime}=\frac{g-g_h}{g} \times 100=\frac{2 h}{R_e} \times 100 \%\)

Effect Of Depth

The gravitational pull on the surface is equal to its weight i.e. mg = \(\frac{G M_e m}{R_e^2}\)

∴ \(\mathrm{mg}=\frac{G \times \frac{4}{3} \pi R_e^3 \rho m}{R_e^2}\)

or \(\mathrm{g}=\frac{4}{3} \pi \mathrm{GR}_{\mathrm{e}} \rho\) ………(1)

When the body is taken to a depth d, the mass of the sphere of radius (R_e – d) will only be effective for the gravitational pull and the outward shall will have no resultant effect on the mass. If the acceleration due to gravity on the surface of the solid sphere is gd, then

⇒ \(g_d=\frac{4}{3} \pi G\left(R_e-d\right) \rho\) ……………(2)

By dividing equation (2) by equation (1)

⇒ \(\mathrm{g}_{\mathrm{d}}=\mathrm{g}\left(1-\frac{d}{R_e}\right)\)

Important Points

At the center of the earth, d = R_e, so g_centre = \(g\left(1-\frac{R_e}{R_e}\right)\) = 0.

Thus, the weight (mg) of the body at the centre of the earth is zero.

Percentage decrease in the value of ‘g’ with the depth

⇒ \(\left(\frac{g-g_d}{g}\right) \times 100=\frac{d}{R_e} \times 100\)

Effect Of The Surface Of Earth:

The equatorial radius is about 21 km longer than its polar radius.

We know, g = \(\frac{G M_e}{R_e^2}\) Hence g_pole > g_equator.

The weight of the body increases as the body is taken from the equator to the pole.

Effect of rotation of the Earth: The earth rotates around its axis with angular velocity ω. Consider a particle of mass m at latitude θ. The angular velocity of the particle is also ω.

⇒ \(\mathrm{g}^{\prime}=\mathrm{g}\left[1-\frac{R_e \omega^2}{g} \cos ^2 \theta\right]\)

At pole θ = 90° ⇒ g_pole = g,

At equator θ = 0 ⇒ g_equator = g \(\left[1-\frac{R_e \omega^2}{g}\right]\)

Hence g _pole > g _equator

If the body is taken from the pole to the equator, then g′ = g\(\left(1-\frac{R_e \omega^2}{g}\right)\)

Hence % change in weight = \(\frac{m g-m g\left(1-\frac{R_e \omega^2}{g}\right)}{m g} \times 100\)

⇒ \(\frac{m R_e \omega^2}{m g} \times 100\)

⇒ \(\frac{R_e \omega^2}{g} \times 100\)

Escape Speed

The minimum speed required to send a body out of the gravity field of a planet (send it to r → ∞)

Escape Speed At Earth’s Surface:

Suppose a particle of mass m is on the earth’s surface

We project it with a velocity V from the earth’s surface so that it just reaches r → ∞ (at r → ∞, its velocity becomes zero)

Applying energy conservation between the initial position (when the particle was at the earth’s surface) and finding positions (when the particle just reaches r → ∞)

K_i+ U_i= K_f+ U_f

⇒ \(\frac{1}{2} m v^2+m_0\left(-\frac{G M_e}{R}\right)\)

⇒ \(0+\mathrm{m}_0\left(-\frac{G M_e}{(r \rightarrow \infty)}\right)\)

⇒ \(\mathrm{v}=\sqrt{\frac{2 G M_0}{R}}\)

Escape speed from earth is surface \(V_e=\sqrt{\frac{2 G M_e}{R}}\)

If we put the values of G, M_e, and R we get

V_e = 11.2 km/s.

Escape Speed Depends On :

1. Mass (Me) and size (R) of the planet
2. Position from where the particle is projected.

Escape Speed Does Not Depend On :

1. Mass of the body which is projected (m0)
2. The angle of projection.

If a body is thrown from the Earth’s surface with escape speed, it goes out of the earth’s gravitational field and never returns to the Earth’s surface. But it starts revolving around the sun.

Kepler’s Law For Planetary Motion

Suppose a planet is revolving around the sun, or a satellite is revolving around the earth, then the planetary motion can be studied with the help of Kepler’s three laws.

Kepler’s Law Of Orbit

Each planet moves around the sun in a circular path or elliptical path with the sun at its focus. (In fact circular path is a subset of an elliptical path)

Law Of Areal Velocity:

To understand this law, let us understand the angular momentum conservation for the planet.

• If a planet moves in an elliptical orbit, the gravitation force acting on it always passes through the centre of the sun. So torque of this gravitation force about the centre of the sun will be zero.
• Hence we can say that the angular momentum of the planet about the centre of the sun will remain conserved (constant) τ about the sun = 0

⇒ \(\frac{d J}{d t}\) = 0

J_planet/ sun = constant

⇒ mvr sinθ = constant

Now we can easily study Kapler’s law of areal velocity.

If a planet moves around the sun, the radius vector \((\vec{r})\) also rotates are sweeps area as shown in the figure. Now let’s find a rate of area swept by the radius vector \((\vec{r})\).

Suppose a planet is revolving around the sun and at any instant its velocity is v, and the angle between radius vector \((\vec{r})\) and velocity \((\vec{v})\). In dt time, it moves by a distance vdt, during this dt time, the area swept by the radius vector will be OAB which can be assumed to be a triangle

dA = 1/2 (Base) (Perpendicular height)

dA = 1/2 (r) (vdtsinθ)

so rate of area swept \(\frac{d A}{d t}=\frac{1}{2} \mathrm{vr} \sin \theta\)

we can write \(\frac{d A}{d t}=\frac{1}{2} \frac{m v r \sin \theta}{m}\)

where mvr sinθ = angular momentum of the planet about the sun, which remains conserved (constant)

⇒ \(\frac{d A}{d t}=\frac{L_{\text {planet } / \text { sun }}}{2 m}\)= constant

so the Rate of area swept by the radius vector is constant

Kepler’s Law Of Time Period: Suppose a planet is revolving around the sun in a circular orbit

then \(\frac{m_0 v^2}{r}=\frac{G M_s m_0}{r^2}\)

v = \(\sqrt{\frac{G M_s}{r}}\)

Time period of the revolution is

⇒ \(\frac{2 \pi r}{v}=2 \pi \mathrm{r} \sqrt{\frac{r}{G M_s}}\)

⇒ \(\mathrm{T}^2=\left(\frac{4 \pi^2}{G M_s}\right) \mathrm{r}^3\)

For all the planets of a sun, T² ∝ r³

Circular Motion Of A Satellite Around A Planet

Suppose a satellite of mass m0 is at a distance r from a planet. If the satellite does not revolve, then due to the gravitational attraction, it may collide with the planet.

To avoid a collision, the satellite revolves around the planet, in a circular motion of a satellite.

⇒ \(\frac{G M_e m_0}{r^2}=\frac{m_0 v^2}{r}\) ….

⇒ v = \(\sqrt{\frac{G M_e}{r}}\) this velocity is called orbital velocity (v0)

⇒ \(\mathrm{v}_0=\sqrt{\frac{G M_e}{r}}\)

Total Energy Of The Satellite Moving In Circular Orbit:

KE = \(\frac{1}{2} m_0 v^2\) and from equation …..(1)

⇒ \(\frac{m_0 v^2}{r}=\frac{G M_e m_0}{r^2}\)

⇒ \(\mathrm{m}_0 \mathrm{v}^2=\frac{G M_e m_0}{r}\)

⇒ \(\mathrm{KE}=\frac{1}{2} m_0 v^2=\frac{G M_e m_0}{2 r}\)

Potential energy

⇒ \(\mathrm{U}=-\frac{G M_e m_0}{r}\)

Total energy = KE + PE = \(\left(\frac{G M_e m_0}{2 r}\right)+\left(\frac{-G M_e m_0}{r}\right)\)

⇒ \(\mathrm{TE}=-\frac{G M_e m_0}{2 r}\)

The total energy is –ve. It shows that the satellite is still bounded by the planet.

Geo – Stationery Satelite

We know that the earth rotates about its axis with angular velocity ω_earth and time period T_earth = 24 hours.

Suppose a satellite is set in an orbit which is in the plane of the equator, whose ω is equal to ω_earth, (or its T is equal to T_earth = 24 hours) and whose direction is also the same as that of earth. Then as seen from Earth, it will appear to be stationary. This type of satellite is called geostationary satellite. For a geo-stationery satellite,

w_satelite = w_earth

⇒ T_satelite = T_earth= 24 hr.

So time period of a geo-stationery satellite must be 24 hours. To achieve T = 24 hours, the orbital radius geo-stationery satellite :

⇒ \(\mathrm{T}^2=\left(\frac{4 \pi^2}{G M_e}\right) \mathrm{r}^3\)

Putting the values, we get an orbital radius of geo stationary satellite r = 6.6 Re(here Re = radius of the earth) height from the surface h = 5.6 Re.

Path Of A Satellite According To Different Speed Of Projection

Suppose a satellite is at a distance r from the centre of the earth. If we give different velocities (v) to the satellite, its path will be different

If v < v₀ \(\left(\text { or } v<\sqrt{\frac{G M_e}{r}}\right)\) then the satellite will move in an elliptical path and strike the earth’s surface.

But if the size of the earth were small, the satellite would complete the elliptical orbit, and the centre of the earth would be at its farther focus.

If v = v₀ \(\left(\text { or } \quad v=\sqrt{\frac{G M_e}{r}}\right)\), then the satellite will revolve in a circular orbit.

If v₀ > V > v₀ \(\left(\text { or } \sqrt{\frac{2 G M_e}{r}}>v>\sqrt{\frac{G M_e}{r}}\right)\), then the satellite will revolve in an elliptical orbital, and the centre of the earth will be at its nearer focus.

If v = v_e \(\left(\begin{array}{ll}\text { or } and v=\sqrt{\frac{2 G M_e}{r}}
\end{array}\right)\), then the satellite will just escape with a parabolic path.

Question 1. Suppose a planet is revolving around the sun in an elliptical path given by \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\). Find a period of revolution. The angular momentum of the planet about the sun is L.

Answer: Rate of area swept = \(\frac{d A}{d t}=\frac{L}{2 m}\) constant

⇒ \(\mathrm{dA}=\frac{L}{2 m} d t\)

⇒ \(\int_{A=0}^{A=\pi a b} d A=\int_{t=0}^{t=T} \frac{L}{2 m} d t\)

⇒ \(\pi \mathrm{ab}=\frac{L}{2 m} \mathrm{~T}\)

⇒ \(\mathrm{T}=\frac{2 \pi m a b}{L}\)

Question 2. The Earth and Jupiter are two planets of the sun. The orbital radius of the earth is 10⁷ m and that of Jupiter is 4 × 10⁷ m. If the time period of the revolution of earth is T = 365 days, find the time period of revolution of the Jupiter.

Answer:

For both the planets T² ∝ r³

⇒ \(\left(\frac{T_{\text {jupitar }}}{T_{\text {earth }}}\right)^2=\left(\frac{T_{\text {jupiter }}}{r_{\text {earhh }}}\right)^3\)

⇒ \(\left(\frac{T_{\text {juppicr }}}{365 \text { days }}\right)^2=\left(\frac{4 \times 10^7}{10^7}\right)^3\)

T_jupiter = 8 × 365 days

Question 3.

Suppose earth has radius R and mass M. A point mass m₀ is at a distance r from the centre. Find the gravitational potential energy of the mass due to the earth.
Answer:

U_g= (m₀) (V_earth)

⇒ \(\mathrm{U}_{\mathrm{g}}=\left(\mathrm{m}_0\right)\left(-\frac{G M_e}{r}\right)=\left(-\frac{G M_e m_0}{r}\right)\)

Question 4.

Suppose the earth has mass and radius R. A small groove in made and point mass m0is placed at the centre of the sphere. With what minimum velocity should we project the particle so that it may escape out of the gravity field (reaches to r → ∞)
Answer:

Suppose the particle is projected with speed v, and to send it to infinity, its velocity should be zero at r → ∞. Applying energy conservation between its initial position (centre) and final position (r → ∞) K_i+ U_i= k_f+ U_f

⇒ \(\frac{1}{2} m_0 v^2+\left(m_0\right)\left(v_{\text {earth }}\right)\)

⇒ \(\frac{1}{2} \mathrm{~m}_0 \mathrm{v}^2+\left(\mathrm{m}_0\right)\left(-\frac{3 G M_e}{2 R}\right)\)

⇒ 0+m₀(0)

⇒ \(\mathrm{v}=\left(\sqrt{\frac{3 G M_e}{R}}\right)\)

Summary

Newton’s Law Of Gravitation:

Gravitational attraction force between two point masses

⇒ \(\mathrm{F}_{\mathrm{g}}=\frac{G m_1 m_2}{r^2}\) and its direction will be attractive.

Gravitational force on (1) due to (2) in vector form

⇒ \(\vec{F}_{12}=\frac{G m_1 m_2}{r^2}\)

Gravitational Field: Gravitational force acting on unit mass.

⇒ \(\mathbf{g}=\frac{F}{m}\)

Gravitational Potential:

⇒ \(\mathrm{v}_{\mathrm{g}}=\frac{U}{m}\)

Gravitation potential energy of unit mass

⇒ \(\mathrm{g}=-\frac{d V_g}{d r}\)

and \(\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}=-\int_A^B \vec{g} \cdot d \vec{r}\)

For point mass: GM

⇒ \(\mathrm{g}=\frac{G M}{r^2}, \mathrm{~V}=-\frac{G M}{r}\)

For circular ring g = \(\frac{G M x}{\left(R^2+x^2\right)^{3 / 2}}\)

⇒ \(\mathrm{v}=-\frac{G M}{\sqrt{R^2+x^2}}\)

For thin circular disc

⇒ \(\mathrm{g}=\frac{2 G M}{R^2}\left(1-\frac{1}{\sqrt{1+\left(\frac{R}{x}\right)^2}}\right)\)

⇒ \(\mathrm{v}=\frac{-2 G M}{R^2}\left(\sqrt{R^2+x^2}-x\right)\)

Uniform thin spherical shell: 

⇒ \(\mathrm{g}_{\text {out }}=\frac{G M}{r^2}\)

⇒ \(\mathrm{g}_{\text {surface }}=\frac{G M}{R^2}\)

g_in = 0

Potential:

⇒ \(\mathrm{v}_{\text {out }}=-\frac{G M}{r}\)

⇒ \(\mathrm{V}_{\text {surtace }}=-\frac{G M}{R}\)

⇒ \(\mathrm{v}_{\mathrm{in}}=-\frac{G M}{R}\)

Uniform solid sphere: (Most Important)

⇒ \(\mathrm{g}_{\text {out }}=\frac{G M}{r^2}\)

⇒ \(\mathrm{g}_{\text {surface }}=\frac{G M}{R^2}\)

⇒ \(\mathrm{g}_{\mathrm{in}}=\frac{G M}{R^3} r\)

⇒ \(\mathrm{g}_{\text {centre }}=0\)

Potential: \(\mathrm{V}_{\text {out }}=-\frac{G M}{r}\)

⇒ \(\mathrm{V}_{\mathrm{in}}=-\frac{G M}{2 R^3}\left(3 R^2-r^2\right)\)

⇒ \(\mathrm{V}_{\text {surtace }}=-\frac{G M}{R}\)

⇒ \(\mathrm{V}_{\text {centre }}=-\frac{3}{2} \frac{G M}{R}\)

Self Energy:

Surface = \(\mathrm{U}_{\text {self }}=-\frac{1}{2} \frac{G M^2}{R}0\)

Gravitational Self energy of a Uniform Sphere = U_self= \(-\frac{3}{5} \frac{G M^2}{R}\)

Escape speed from earth’s surface

⇒ \(\mathrm{V}_{\mathrm{e}}=\sqrt{\frac{2 G M_e}{R}}\)

= 11.2km/sec.

If a satellite is moving around the Earth in a circular orbit, then its orbital speed is

⇒ \(\mathrm{V}_0=\sqrt{\frac{G M_e}{r}}\)

where r is the distance of the satellite from the centre of the earth.

PE . of the satellite = – \(\frac{G M_e m}{r}\)

KE of the satellite = \(\frac{1}{2} m v_0^2=\frac{G M_e m}{2 r}\)

TE of the satellite = \(-\frac{G M_e m}{2 r}\)

Time Period of Geo-stationary satellite = 24 hours

Kapler’s laws: 

1. Law of Orbit: If a planet is revolving around a sun, its path is either elliptical (or circular)
2. Law of Area :

View (1) If a planet is revolving around a sun, the angular momentum of the planet about the sun remains conserved

View (2) The radius vector from the sum to the planet sweeps the area at a constant rate

Areal velocity = \(\frac{d A}{d t}=\frac{L}{2 m}\) = constant

(3) For all the planets of a sun T² ∝ R³

⇒ \(\mathrm{T}^2=\left(\frac{4 \pi^2}{G M_s}\right) \mathrm{R}^3\)

Factors Affecting Acceleration Due to Gravity

1. Effect Of Altitude: \(\mathrm{g}_{\mathrm{n}}=\frac{G M_e}{\left(R_e+h\right)^2}\)

\(\mathrm{g}\left(1+\frac{h}{R_e}\right)^{-2} \simeq \mathrm{g}\left(1-\frac{2 h}{R_e}\right)\) when h << R.

2. Effect Of Depth: \(\mathrm{g}_{\mathrm{d}}=\mathrm{g}\left(1-\frac{d}{R_e}\right)\)

3. Effect Of The Surface Of Earth

The equatorial radius is about 21 km longer than its polar radius.

We know, g = \(\frac{G M_e}{R_e^2}\)

Hence g_pole > g_equator.

4. Effect Of Rotation Of The Earth

Consider a particle of mass m at latitude θ. g′ = g – ω²R_e cos²θ

At pole θ = 90°

⇒ g_pole = g , At equator θ = 0

⇒ g_equator = g – ω²Re.

Hence g_pole > g_equator