By going through these CBSE Class 11 Physics Notes Chapter 8 Gravitation, students can recall all the concepts quickly.

## Gravitation Notes Class 11 Physics Chapter 8

→ Gravitation is a central force.

→ It acts along the line joining the particles.

→ Gravitation is the weakest force in nature.

→ It is about 10^{38} times smaller than the nuclear force and 10^{36} times smaller than the electric force.

→ Gravitation is the conservative force.

→ Gravitation is caused by gravitational mass.

→ Gravitation acts in accordance with Newton’s third law of motion. That is F_{12} = – F_{21}.

→ Gravitation is independent of the presence of the other bodies in the surroundings.

→ Acceleration due to gravity is 9.81 ms^{-2} on the surface of Earth.

→ Its value on the moon is about one-sixth of that on Earth.

→ Its value on Sun, Jupiter and Mercury is about 27 times, 2.5 times and 0.4 times that on the Earth. i.e. g_{moon} = g_{e}/6. g_{sun} = 27g, g_{mercury} = 0.4g, g_{jupiter} = 2.5g.

→ The value of g (acceleration due to gravity) does not depend upon the mass, shape or size of the falling body.

→ Inside the Earth, the value of g decreases linearly with distance from the centre of Earth.

→ Above the surface of Earth, the value of g varies inversely as the square of the distance from the centre of Earth.

→ The value of g decreases faster with altitude than with depth.

→ For small values of height (h), the value of g at a height ‘h’ is the same as the value of g at a depth d (= 2h).

→ Decrease in g at a height h = x (very near the Earth’s surface i.e. h << R) is twice as compared to the decrease in g at the same depth d = x.

→ g is maximum on Earth’s surface and decreases both when we go above or below the Earth’s surface.

→ The value of g is zero at the centre of Earth.

→ The rate of variation of g with height (near the surface of Earth, when h << R) is twice the rate of variation of g with depth i.e.

\(\frac{\Delta \mathrm{g}_{\mathrm{h}}}{\Delta \mathrm{h}}\) = 2 \(\frac{\Delta \mathrm{g}_{\mathrm{d}}}{\Delta \mathrm{d}}\)

→ With the increase in latitude, g decreases.

→ Its value at latitude Φ is given by

g_{Φ} = g_{p} – Rω^{2} cos^{2}Φ

→ The decrease in ‘g’ with latitude is due to the rotation of Earth about its own axis.

→ The decrease in ‘g’ with latitude on rotation is because a part of the weight is used to provide centripetal force for the bodies rotating with the Earth.

→ The g is maximum at poles (g_{p}) and minimum at the equator (g_{e}).

→ g_{p} = 9.81 ms^{-2}, g_{c} = 9.78 m^{-2}.

→ The decrease in g from pole to equator is about 0.35%.

→ If the Earth stops rotating about its own axis, the value of g on the poles will remain unchanged but at the equator, it will increase by about 0.35%. If the rotational speed of Earth increases, the value of g decreases at all places on its surface except poles.

→ The gravitational pull of Earth is called true weight (w_{t}) of the body i.e. W_{t} = mg.

→ The true weight of the body varies in the same manner as the /acceleration due to gravity i.e. it decreases with height above and depth below the Earth’s surface.

→ Also, Wt changes with latitude. Its value is maximum at the poles and minimum at the equator.

→ S.I. unit of weight is Newton (N) and it is often expressed in kilogram weight (kg wt) or kg f (kilogram-force).

i. e. 1 kg wt = 1 kg f = 9.8 N

→ The reaction of the surface on which a body lies is called apparent weight (W_{a}) of the body and ga = W_{a}/M is called apparent acceleration due to gravity.

→ If a body moves with acceleration a, then the apparent weight of the body of mass M is given by W_{a} = M|g – a| = apparent weight of the body when it falls with acceleration ‘a’ and it decreases.

→ When a body rises with acceleration, then W_{a} = M(g + a) i.e. it increases.

→ If the body is at rest or moving with uniform velocity, then W_{a} = Mg = true weight of the body.

→ For free-falling body, W_{a} = 0 (∵ a = g here).

→ The spring balance measures the apparent weight of the body.

→ The apparent weight provides restoring force to the simple pendulum i.e. the time period of the simple pendulum depends on the apparent value of the acceleration due to gravity i.e.

T = 2π\(\sqrt{\frac{l}{\mathrm{~g}_{\mathrm{a}}}}\)

→ In a freely falling system, the time period of a simple pendulum is infinity.

→ If a simple pendulum is suspended from the roof of an accelerating or retarding train, the time period is given by

T = 2π\(\sqrt{g^{2}+a^{2}}\)

→ The Earth is ellipsoid. It is flat at poles and bulges out at the equator. Consequently, the distance of the surface from the centre is more at the equator than on the poles. So g_{pole} > g_{equator}.

→ All bodies fall freely with the same acceleration = g.

→ The acceleration of the falling body does not depend on its mass.

→ If two bodies are dropped from the same height, they reach the ground at the same time with the same velocity.

→ If a body is thrown upward with velocity u from the top of a tower and another is thrown downward from the same point and with the same velocity, then both reach the ground with the same speed.

→ If a body is dropped from a height h, it reaches the ground with speed v = \(\sqrt{2 \mathrm{gh}}\) = gt. Time taken by it to reach the ground is t = \(\sqrt{\frac{2 h}{g}}\)

→ When a body is dropped, then initial velocity i.e. u = 0.

→ If a body is dropped from a certain height h, then the distance covered by it in nth second is \(\frac{1}{2}\) g(2n -1).

→ The greater the height of a satellite, the smaller is the orbital velocity. Work done to keep the satellite in orbit is zero.

→ When a body is at rest w.r.t. the Earth, its weight equals gravity and is known as its true or static weight.

→ The centripetal acceleration of the satellite = acceleration due to gravity.

→ Orbital velocity is independent of the mass of the satellite.

→ Orbital velocity depends on the mass of the planet as well as the radius of the orbit.

→ All communication satellites are geostationary satellites.

→ Escape velocity (V_{e}) from the surface of Earth = 11.2 km s^{-1}.

→ The body does not return to the Earth when fired with V_{e} irrespective of the angle of projection.

→ When the velocity of the satellite increases, its kinetic energy increases and hence total energy becomes less negative i.e. the satellite begins to revolve in orbit of greater radius.

→ If the total energy of the satellite becomes +ve, the satellite escapes from the gravitational pull of the Earth.

→ When the satellite is taken to a greater height, the potential energy increases (becomes less negative) and the K.E. decreases.

→ For the orbiting satellite, the K.E. is less than the potential energy. When K.E. = P.E., the satellite escapes away from the gravitational pull of the Earth.

→ The escape velocity from the moon is 2.4 km s^{-1}.

→ The ratio of the inertial mass to gravitational mass is one.

→ If the radius of a planet decreases by n% keeping mass constant, then g on its surface decreases by 2n%.

→ If the mass of a planet increases by m% keeping the radius constant, then g on its surface increases by m%.

→ If the density of the planet decreases by x% keeping the radius constant, the acceleration due to gravity decreases by x%.

→ The intensity of the gravitational field inside a shell is zero.

→ The weight of a body in a spherical cavity concentric with the Earth is zero.

→ Gravity holds the atmosphere around the Earth.

→ The reference frame attached to Earth is non-inertial because the Earth revolves about its own axis as well as about the Sun.

→ When a projectile is fired with a velocity less than the escape velocity, the sum of its gravitational potential energy and kinetic energy is negative.

→ If the Earth were at one fourth the present distance from Sun, the duration of the year will be one-eighth of the present year.

→ The tail of the comets points away from the Sun due to the radiation pressure of the Sun.

→ Even when the orbit of the satellite is elliptical, its plane of rotation passes through the centre of the Earth.

→ If a packet is just released from an artificial satellite, it does not fill the Earth. On the other hand, it will continue to orbit with the satellite.

→ Astronauts orbiting around the Earth cannot use a pendulum clock.

→ However, they can use a spring clock.

→ To the astronauts in space, the sky appears black due to the absence of the atmosphere above them.

→ The duration of the day from the moment the Sun is overhead today to the moment the Sun is overhead tomorrow is determined by the revolution of the Earth about its own axis as well as around the Sun.

→ If the ratio of the radii of two planets is r and the ratio of the acceleration due to gravity on their surface is ‘a’ then the ratio of escape velocities is \(\sqrt{ar}\).

→ Two satellites are orbiting in circular orbits of radii R_{1} and R_{2}. Their orbital speeds are in the ratio: \(\frac{V_{1}}{V_{2}}=\left(\frac{R_{2}}{R_{1}}\right)^{\frac{1}{2}}\). It is independent of their masses.

→ An object will experience weightlessness at the equator if the angular speed of the Earth about its axis becomes more than (\(\frac{1}{800}\)) rad/s.

→ If a body is orbiting around the Earth then it will escape away if its velocity is increased by 41.8% or when its K.E. is doubled.

→ If the radius of Earth is doubled keeping mass unchanged, the escape \(\left(\frac{1}{\sqrt{2}}\right)\) times the present value.

→ V_{o} close to Earth’s surface = 7.9 km s^{-1}.

→ The time period of the satellite very near the surface of Earth is about 107 minutes.

→ No energy is dissipated in keeping the satellite in orbit around a planet.

→ If a body falls freely from infinite height, then it reaches the surface of Earth with a velocity of 11.2 km s^{-1}.

→ A body in a gravitational field has maximum binding energy when it is at rest.

→ Acceleration due to gravity: The acceleration with which bodies fall towards the Earth is called acceleration due to gravity.

→ Newton’s law of gravitation: It states that everybody in this universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres i.e.

F = \(\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\)

→ Gravitation: Force of attraction between any two bodies.

→ Gravity: Force of attraction between Earth and any other body.

→ Inertial mass: The resistance to the acceleration caused by a force is called inertial mass. It is the measure of its inertia in linear motion

i.e. m1 = \(\frac{\mathrm{F}}{\mathrm{a}}\)

→ Gravitational mass: The resistance to the acceleration caused by gravitational force i.e.

mg = \(\frac{\mathrm{F}}{\mathrm{g}}\)

→ Kepler gave threads of planetary motion.

→ Kepler’s 1st law of planetary motion: Every planet revolves around Sun in an elliptical orbit with Sun at one of its foci.

→ Second law: The radius vector joining the centre of Sun and planet sweeps out equal areas in equal intervals of time i.e. areal velocity of the planet around the Sun always remain constant.

→ Third law: The square of the time period T of revolution of a planet around the Sun is proportional to the cube of the semi-major axis R of its elliptical orbit i.e.

T^{2} ∝ R^{3}

→ Satellite: It is a body that constantly revolves in an orbit around a body of relatively much larger size.

→ A geostationary satellite is a satellite that appears stationary to the observer on Earth. It is also called a geosynchronous satellite.

- Its orbit is circular and in the equatorial plane of Earth.
- Its time period = time period of rotation of the Earth about its own axis i.e. one day or 24h = 86400s.
- Its height is 36000 km.
- Its orbital velocity is about 3.08 km s
^{-1}. - Its angular velocity is equal and is in the same direction as that of Earth about its own axis.

→ Latitude at a place: Latitude at a place on the Earth’s surface is the angle at which the line joining the place to the centre of Earth makes with the equatorial plane. It is denoted by X.

→ At poles, X = 90°.

→ At equator, X = 0.

→ Polar satellites: They are positioned nearly 450 miles above the Earth. Polar satellites travel from pole to pole in nearly 102 minutes.

→ In each successive orbit, the satellite scans a strip of the area towards the West.

→ Orbital Velocity: It is the velocity required to put the satellite in a given orbit around a planet. It is denoted by V0.

→ Escape Velocity: It is the minimum velocity with which a body be thrown upwards so that it may just escape the gravitational pull of Earth or a given planet. It is denoted by Ve.

→ The intensity of Gravitational field at a point: It is the force experienced by a unit mass placed at that point.

→ The gravitational potential energy of a body at a point in a gravitational field is the amount of work done in bringing the body from infinity to that point without acceleration.

→ Gravitational Potential: It is the amount of work done in bringing a body of unit mass from infinity to that point without acceleration.

**Important Formulae:**

→ The universal force of gravitation,

F = \(\frac{\mathrm{Gm}_{1} \mathrm{~m}_{2}}{\mathrm{r}^{2}}\)

→ Invector from F = \(\frac{\mathrm{Gm}_{1} m_{2} \hat{r}}{r^{3}}\)

→ G = 6.67 × 10^{-11} Nm^{2} kg^{-2}.

→ Weight of the body, W = mg.

→ Mass of Earth, Me = \(\frac{\mathrm{g} \mathrm{R}_{\mathrm{e}}^{2}}{\mathrm{G}}\)

→ Gravitational mass, mg = \(\frac{\mathrm{F}}{\mathrm{g}}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}_{\mathrm{g}}}{\mathrm{g} \mathrm{R}_{\mathrm{e}}^{2}}\)

→ Variation of g:

- At height h: gh = g\(\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}\) ≈ (1 – \(\frac{2 \mathrm{~h}}{\mathrm{R}}\))
- At depth d: g
_{d}= g(1 – \(\frac{\mathrm{d}}{\mathrm{R}}\)) - At latitude λ: g
_{λ}= g – Rω^{2}cos^{2}λ

(a)At poles: λ = 90°, cosλ = 0 ∴ g_{λpole}= g

(b) At equator: λ = 0, cosλ = 1, ∴ g_{λ}= g Rω^{2}

→ g_{pole}– g_{equator} = Rω^{2}

→ Gravitation field intensity: (I) = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)

or

|I| = g

→ Orbital velocity in orbit at a height h,

v_{o} = \(\sqrt{\frac{\mathrm{gR}^{2}}{\mathrm{R}+\mathrm{h}}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\mathrm{h}}}\)

→ If h = 0 i.e. close to Earth’s surface, v_{o} = \(\sqrt{gR}\).

→ Time period of the satellite,

T = \(\frac{2 \pi(\mathrm{R}+\mathrm{h})}{\mathrm{v}_{0}}=\frac{2 \pi}{\mathrm{R}} \sqrt{\frac{(\mathrm{R}+\mathrm{h})^{3}}{\mathrm{~g}}}\)

→ Escape velocity, V_{e} = \(\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}=\sqrt{2 \mathrm{gR}}=\sqrt{2}\)V_{o}

= \(\sqrt{\frac{8}{3} \pi R^{3} G \rho}\)

→ Gravitational potential energy,

E = – \(\frac{\mathrm{GMm}}{\mathrm{r}}\)

→ Self energy of Earth = – \(\frac{3}{5} \frac{\mathrm{GM}^{2}}{\mathrm{R}}\)

→ Increase in gravitational RE. when the body is moved from surface of Earth to a height h,

ΔE = E_{h} – E_{e}

= – \(\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}-\frac{\mathrm{GMm}}{\mathrm{R}}\)

= \(\frac{\text { GMmh }}{R(R+h)}\)

→ Gravitational potential, V = – \(\frac{\mathrm{GM}}{\mathrm{r}}\)

→ Time period of motion of the satellite:

T^{2} = \(\frac{4 \pi^{2}}{\mathrm{GM}_{\mathrm{E}}}\) r^{3} = \(\frac{4 \pi^{2}}{\mathrm{gR}_{\mathrm{E}}^{2}}\) r^{3}

→ Angular velocity (ω) of a satellite in an orbit at a height h above Earth.

ω = \(\frac{2 \pi}{T}=\sqrt{\frac{G M}{(R+h)^{3}}}=\sqrt{\frac{g_{h}}{R+h}}\)

→ Shape of the orbit of a satellite having velocity v in the orbit:

- If v < v
_{o}, the satellite falls to the Earth following a spiral path. - If v = v
_{o}, the satellite continues to move in orbit. - If v
_{o}< v < v_{es}, then the satellite moves in an elliptical orbit, - If v = v
_{es}, then it escapes from Earth following a parabolic path, - If v > v
_{es}, then the satellite “escapes from Earth following a hyperbolic path.