Gravitation |

 Newton’s law of Gravitation:

Everybody in the universe attracts other bodies towards its centre. The force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of distance between their centers

$F \propto \dfrac{m_1 m_2}{r^2}$

$F=G \dfrac{m_1 m_2}{r^2}$

Where $G=6.67×10^{-11} Nm^2/kg^2$ is called universal gravitational constant whose value remain same at all the places and for all bodies but depends on system of units.

$\textbf{Gravitational Force:}$

(1) Always attractive, conservative and central force depending upon the masses of the bodies but is independent of medium and presence of any other bodies nearby.

(2) It acts along the line joining the centers of the bodies.

(3) It produces action and reaction and hence obeys Newton’s third law of motion.

(4) It is effective for large distance (Nuclear force is effective only for some very small distance.

(5) Comparison of strength :

It is weaker than nuclear, electric and magnetic forces but stronger than intermolecular force.

(6) It exists between any pair of bodies throughout the universe.

Consequence of gravitational force being a central force: Angular momentum of any heavenly body is constant as torque becomes zero relative to sun.

$\textbf{Weight}$

The weight of a body is the total gravitational force exerted on the body by all other bodies in the universe. When the body is near the surface of the earth, we can neglect all other gravitational forces and consider the weight as just the earth’s gravitational attraction. At the surface of the moon we consider a body’s weight to be the gravitational attraction of the moon, and so on.

It is a vector quantity. It is always directed towards the centre of the planet.

It holds the atmosphere around the earth.

Mathematically,

W = mg

Where m = mass of the body

g = acceleration due to gravity on the surface of that heavenly body

for earth g = 9.8$m/s^2$

$\textbf{Acceleration due to gravity (g)}$

It is the acceleration produced by Earth or planets towards centre of itself.

It is a vector quantity.

According to Newton’s law of motion ( 2nd law), force of gravity is F=mg………….(1)

And we have, $F=\dfrac{GMm}{r^2}$

Where m is the mass of the body

So, comparing we get, $g=\dfrac{GM}{r^2}$

If a body is at the surface of earth, then

$g=\dfrac{GM}{R^2}$ Where r=R is the radius of this earth

M is the mass of planet or earth

In terms of density:

Also $g=\dfrac{4}{3} \pi G \rho R$

Where $\rho$ is the density of earth/planet The value of g does not depend upon mass size and shape of the body.

Variation of g:

$\text{Due to shape of the earth:}$

Equatorial radius of the earth is about 21 km more than the polar radius so, according to $g \propto \dfrac{1}{R^2}$ , value of g at poles is greater by $0.018m/s^2$ $(1.8 cm/s^2)$ than that at the equator.

$\textbf{Due to rotational motion of the earth:}$

If the earth rotates about an axis with angular velocity(ω), then effective g on a body at latitude θ is given by:

$g_{eff} =g'-Rω^2 cos^2⁡θ$

• At equator; i.e. $θ=0^{\circ}$

∴$g_{eff}=g-Rω^2$

$Δg=g-g_{eff} =Rω^2=0.034m/s^2$

• At poles,

$θ=90^{\circ}$

i.e $g_{eff} =g-0=g$ (i.e. No rotational effect at poles)

$\textbf{Due to height above the earth’s surface:}$

Acceleration due to gravity at a height from the earth’s surface is given as:

Acceleration due to gravity at a height from earth's surface is given by:

$g' = \dfrac{GM}{(R+h)^2}$

where, h = height of the body above the earth's surface

If $h <<< R$ then, $g' = g (1 - \dfrac{2h}{R})$

$\textbf{Due to depth (x) below earth surface:}$

Acceleration due to gravity (g) below depth (x) from earth surface is given by:

$g = \dfrac{GM}{R^3} (R-x) = g (1- \dfrac{x}{R}) = \dfrac{g}{R} (R-x)$

The value of g decreases with increase in depth.

At center of earth, $x=R, g=0$

Gravitational Field Strength/ Gravitational Intensity (E)

Gravitational intensity at a point in a gravitational field is force experienced by a unit point mass when placed at that point.

It shows how much a body is attracted by the force. Gravitational field due to the earth at a place on the surface of the earth is constant for any object be it an ant or an elephant. It depends on the mass of the heavenly body (Earth) but not on the mass of body experiencing the force.

$E=F/m= \frac{GM}{r^2}$

-Its SI unit is $Nkg^{-1}$

-It is a vector quantity.

-Its value is zero at infinity.

Gravitational Potential Energy:

Gravitational potential energy is the energy possessed or acquired by an object due to a change in its position when it is present in a gravitational field. In simple terms, it can be said that gravitational potential energy is an energy that is related to gravitational force or to gravity.

The most common example that can help you understand the concept of gravitational potential energy is if you take two pencils. One is placed at the table and the other is held above the table. Now, we can state that the pencil which is high will have greater gravitational potential energy that the pencil that is at the table.

When a body of mass (m) is moved from infinity to a point inside the gravitational influence of a source mass without accelerating it, the amount of work done in displacing it into the source field is stored in the form of potential energy this is known as gravitational potential energy. It is represented with the symbol (U).

We know that the potential energy of a body at a given position is defined as the energy stored in the body at that position. If the position of the body changes due to the application of external forces the change in potential energy is equal to the amount of work done on the body by the forces. The gravitational force is conservative: The work done by gravitational force does not depend on the path taken from the earth’s surface to a certain height be it curved or straight path. The gravitational force on a body at infinity is zero; therefore, potential energy is zero, which is called a reference point.

$U=-\dfrac{GMm}{r}$

Where r = distance of the body from the centre of the earth

At the surface of the earth:

$U=- \dfrac{GMm}{R}$ ; R = Radius of the earth

The change in gravitational potential energy ΔU when that body is raised through height h from earth surface is given as

$ΔU=U-U_0=mgh(\dfrac{R}{R+h})$

For $h<<<R, \dfrac{R}{R+h}=1$

$∴ΔU=mgh$

Note: The famous formula of Potential energy i.e. P.E. = mgh that we learnt during school days is change in P.E. of a body where height changed is very small in comparison to the Radius of the earth.

Gravitational Potential:

It is the work done to take a unit mass from infinity to a certain point of consideration within the gravitational field.
$V=- \dfrac{GM}{r}$
It is a scalar quantity.

Relations between gravitational force (f), intensity (E), potential (V) and gravitational potential energy (U)

(i) Gravitational force F

$F=-dU/dr=mE= \dfrac{GMm}{r^2} =-m \dfrac{dV}{dr}$

(ii) Gravitational intensity E

$E=-dv/dr=F/m= \dfrac{GM}{r^2} = \dfrac{-1}{m} \dfrac{dU}{dr}$

(iii) Gravitational potential V

$V=- \int Edr= \dfrac{-1}{m}∫Fdr=U/m= \dfrac{-GM}{r}$

(iv) Gravitational potential energy U

$U=- \int Fdr=-m \int Edr=mV= \dfrac{-GMm}{r}$

Motion Of satellite:

1. Speed of a Satellite

$\dfrac{mv_0^2}{r} = \dfrac{GMm}{r^2}$

$v_0 = \sqrt{\dfrac{GM}{r}} = \sqrt{\dfrac{GM}{R+h}} = \sqrt{\dfrac{gR^2}{R+h}}$
2. Angular velocity of satellite
$ω \dfrac{v_0}{r}=\sqrt{\dfrac{GM}{r^3}}$
Where r =R (When it is close to earth surface)
$r=R⇒ω=\sqrt{g/R}=\dfrac{1}{800} rad/s$
3. Time period of satellite
$T = \dfrac{2\pi}{\omega} = 2\pi \sqrt{\dfrac{r^3}{GM}}$
$T^2 \propto r^3$
At r= R, $T = 2 \pi \sqrt{\dfrac{R}{g}} = 84.6 min$
4. Kinetic energy (KE) of a satellite in orbit
$K.E.= \dfrac{1}{2}mv_0^2= \dfrac{GMm}{2r}= \dfrac{mgR^2}{2(R+h)}$
5. Potential energy (U) of a satellite in orbit
$U= \dfrac{-GMm}{r}=-\dfrac{mgR^2}{r}=-2K.E.$
6. Total energy (E.) is given by

$E=P.E.+ K.E.= \dfrac{-GMm}{2r}=-K.E.$

[Note: Binding energy of a satellite is the energy required by it in orbit just to escape which is given by $B⋅E.= \dfrac{GMm}{2r}=K.E.=-T.E.$]

Geostationary satellites:

Those satellites whose period of revolution is equal to the time period of rotation of earth are geostationary satellites.

These satellites appear to be stationary to a person on earth surface. The orbit of these satellites are called parking orbits. The orbital velocity and radius of orbit of geostationary satellites are 3.1km/sec and 42400km.

It’s height is nearly 36000km. If r is the radius of parking orbit then

$m\omega^2 r = \dfrac{GMm}{r^2}$

$r = (\dfrac{GMT^2}{4\pi^2})^{1/3}$

Here, $g = 9.8 m/s^2, R = 6.4 \times 10^6 m, T = 24 hr$ then $r = 42400$ km

Orbital velocity of parking obit is given by

$v_0= \sqrt{\dfrac{GM}{r}}=3.1km/s$

Escape velocity:

Minimum velocity imparted to a body at a point in gravitational field of earth planet so that it just escapes the gravitational field of earth/planet is called escape velocity.

When a body is projected with the energy equal to its gravitational potential energy, then it acquires escape velocity

$v_e = \sqrt{\dfrac{2GM}{R}}= \sqrt{2gR}$

$v_e$ depends on mass, density, radius, height of planet, acceleration due to gravity of the planet but not on the mass density shape and size of the body to be projected.

$v_e$ does not depend on direction of projection.

If a body is projected from a certain height h above the earth’s surface then,

$\dfrac{GMm}{R+h} = \dfrac{1}{2} m v_e^2$

$v_e = \sqrt{\dfrac{2GM}{r}} = \sqrt{\dfrac{2GM}{R+h}}$

From earth surface, h=0,r=R

∴$v_e =\sqrt{2gR} =11.2km/s$

Keplar’s Laws:

$\text{1st Law:}$

According to Kepler’s first law “All the planets revolve around the sun in elliptical orbits having the sun at one of the foci”. The point at which the planet is close to the sun is known as perihelion and the point at which the planet is farther from the sun is known as aphelion.

It is the characteristics of an ellipse that the sum of the distances of any planet from two foci is constant. The elliptical orbit of a planet is responsible for the occurrence of seasons.

$\textbf{2nd law:}$

Kepler’s second law states “The radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time.”

As the orbit is not circular, the planet’s kinetic energy is not constant in its path. It has more kinetic energy near perihelion and less kinetic energy near aphelion implies more speed at perihelion and less speed $(v_{min})$ at aphelion. If r is the distance of planet from sun, at perihelion $(r_{min})$ and at aphelion $(r_{max})$, then,

$r_{min} + r_{max}$= 2a × (length of major axis of an ellipse) . . . . . . . (1)

This law is based on conservation angular momentum.

According to Kepler’s 2nd law, the speed of the planet is maximum when it is closest to the sun and is minimum when the planet is farthest from the sun.

$\textbf{3rd law}$

According to Kepler’s law of periods," The Square of the time period of revolution of a planet around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis".

$T^2 \propto a^3$

Shorter the orbit of the planet around the sun, shorter the time taken to complete one revolution. Using the equations of Newton’s law of gravitation and laws of motion, Kepler’s third law takes a more general form:

$P^2= \dfrac{4π^2}{[G(M_1+M_2 )]} \times a^3$

where $M_1$ and $M_2$ are the masses of the two orbiting objects in solar masses.

It means larger is the distance of a planet from the sun, greater will be its time period of revolution of nearest.

According to Kepler’s 3nd law, as the distance of the planet from sun increases, duration of the year of planet increases.

The planets which are orbiting around the sun their angular speed, linear speed, K.E. change with time but angular momentum remains constant.

Gravitational and Inertial Mass

• Gravitational Mass

• The ratio of weight of a body to acceleration due to gravity is called gravitational mass.

$M_g= \dfrac{FR^2}{GM}=F/g$

• It is affected by the presence of other bodies nearby and it is measured with spring balance or physical balance.
2. Inertial Mass
• The ratio of force applied on a body to acceleration produced is called inertial mass.
$m_i=F/a$
• It is not affected by the presence of other bodies nearby. It is proportional to quantity of matter present in the body.

• It does not depend on shape, size and state of body Inertial mass increases as speed of the body increases according to relation.
$M =\dfrac{m_0}{\sqrt{1-\dfrac{v^2}{c^2}}}$
• It is measured by inertial balance. [Note: Gravitational mass and inertial mass are equivalent. But their definitions are independent of one another.]

Better You Know

At the same point, $v_e:v_0= \sqrt{2}:1$

If speed of satellite increases by 41.4% or its K.E increases by 100% then it will escape to infinity.

Since the satellite has total negative energy, it is bound to revolve in the orbit.

When a satellite is orbiting in its orbit, zero energy is spent to keep it in its orbit.

Air friction increases the velocity of the satellite.

Gravitational mass is proportional to gravitation force.

Orbital velocity depends on mass, radius, density and acceleration due to gravity of earth and on the height (h) of the satellite.

A body weights very slightly more at night since gravitational attraction of moon also adds at night

If polar ice caps melt, then moment of inertia (I) and time period of rotation of earth increase.

Spring balance measures apparent weight, physical balance measures gravitational mass.

A body weights very slightly more at night since the gravitational attraction of moon also adds at night.

A body orbiting the earth will escape its pull when

i) its K.E is doubled i.e ↑by 100% .

ii) its velocity is increased by 41.8 % .

Spring balance measures apparent weight, physical balance measures gravitational mass.

Gravitational mass is proportional to gravitational force.

If polar ice caps melts, then a moment of inertia (i) and time period of rotation of earth increase.

Two air bubble is inside water attracts each other.

$F = \dfrac{Gm_1 m_2}{r^2} = \dfrac{G(-m_1) (-m_2)}{r^2} = \dfrac{Gm_1 m_2}{r^2}$

An air bubble is repelled from a metal sphere

$F = \dfrac{Gm_1 m_2}{r^2} = \dfrac{G(-m_1) (m_2)}{r^2} = \dfrac{- Gm_1 m_2}{r^2}$

Mass of air bubble in the material medium is negative.

When a satellite is orbiting in its orbit, zero energy is spent to keep it in its orbit.

The change in Gravitational PE When a body is raised through height is

$∆U=U-U_0= \dfrac{GMmh}{R(R+h)}=mgh \dfrac{R}{(R+h)}$

But the small heights i.e R>>> h → $\dfrac{R}{(R+h)} ≈1$

$∆U=mgh$

When a body is projected horizontally with velocity v from any other heights from the surface of earth , then the following possibilities are there ,

i) If $v < v_0$, the body fails to revolve around earth and finally falls to the surface of the earth .

ii) If $v = v_0$, then the body will revolve around the earth in circular orbit .

iii) If $v < v_e$, then the body will revolve around the earth in elliptical orbit .

iv) If $v = v_e$, the body will escape from the gravitational field of earth .

v) If $v > v_e$, then the body will escape following a hyperbolic path .

If earth rotates at an angular velocity $1.25 \times 10^{-3}$ rad/sec or 1/800 rad/sec i.e 17 times the present value then particle become weightless at the equator (g =0 ) : But the value of g is unchanged at poles. In such case, duration of the day will be 1 hr and 14 min ( 84.6 mins).

If a body is imparted a velocity y from earth surface, then

i) If $v < v_e$ then T .E ( = P.E + K.E ) < 0 . i.e –ve ; the body returns to earth after attaining the maximum height h given by $h = \dfrac{R}{(\dfrac{2gR}{v^2} -1)}= \dfrac{R}{(\dfrac{v_c^2}{v^2} -1)}$

ii) If $v = v_e$, then T.E zero ; body just escapes to infinity with zero velocity .

iii) If $v > v_e$ then T.E > 0 i.e +ve body escapes infinity with certain speed given by : $v'= \sqrt{(n^2-1)} v_e$

where $v = nv_e$

The hydrogen balloon released on the moon would fall with an acceleration (9.8/6) m/s$^2$ ( i.e 1.6m/s$^2$ ) since there is no air to impart upthrust .

If a body of density $ρ_2$ is falling under gravity in a medium of density $ρ_1$ then the effective value of acceleration due to gravity is given by,

$g’ = g(1 - \dfrac{ρ_1}{ρ_2})$

If two bodies of different masses are dropped simultaneously from the same height in air, the heavier body will reach the ground first.

In a freely falling body, time period of simple pendulum is infinity.

For a freely falling body, apparent weight is zero.

If two planets have radius $r_1$ and $r_2$ and densities $d_1$ and $d_2$ respectively , then ratio of acceleration due to gravity on them will be $g_1:g_2 = r_1d_1 : r_2d_2$.

If a body is released from an artificial satellite it doesn’t fall to the earth but will continue orbiting along with the satellite .

The ratio of energy required to raised a satellite to height h above the earth’s surface to that required to put it into the orbit is 2h:R .

A Spaceship moves from earth to moon and back . The greatest energy required for the spaceship is to overcome the difficulty in take off from the earth’s field .

Work done to keep the satellite in orbit is zero .

When velocity of satellite increases , K.E increases and T.E becomes less –ve . Thus, satellite begins to revolve in orbit greater radius .

When satellite is taken to greater height , P.E . increases and K.E decreases .

For orbiting satellite , KE < PE .

When KE = PE , satellite escapes away from gravitational pull of earth .

For an orbiting satellite ,

K.E = GMm/2r ,P.E= -GMm/r ,T.E= -Gmm/2r

K.E = -T.E

P.E = 2K.E

P.E = 2T.E

The value of acceleration due to gravity becomes 1/n times its value at earth’s surface at

i) distance $r = R/n$ from center of the earth.

ii) depth $x = \dfrac{n-1}{n} R$ from earth's surface

If the earth suddenly stops rotating about its axis , the value of g at the equator will increase by $ώ^2R$. [HINT : $g_e = g- Rώ^2$ ( at equator ) and at poles no rotational effect ]

The time period of geostationary satellite is 24 hours ( that of earth )

The height of communication satellite from earth’s surface is 36000 km .

f two masses $M_1$ and $M_2$ having radii $R_1$ and $R_2$ are d distance apart then minimum velocity with which a particle of mass m should be projected from the midpoint between the centers of bodies is given by $v = 2 \sqrt{ \dfrac{G ( M_1+ M_2)}{d}}$

Two particles of equal masses ( each of mass m ) go around a circle of radius R under the action of their mutual attraction, then the speed of each particle is given by v = 1/2 $\sqrt{\dfrac{GM}{R}}$

The value of acceleration due to gravity becomes 1/n times its value on surface of the earth at i) distance $r = \sqrt{n} R$ from center of earth .

ii) height $h = (\sqrt{n}-1 )R$ from earth’s surface .

where R is the radius of earth ..

Weightlessness experienced while orbiting the earth in a spaceship is the result of acceleration .

The escape velocity of a body does not depend on its mass but depends on the mass of the planet .

Two heavenly bodies not far off from each other are seen to revolve around their common center of mass .

If a piece of rock falls vertically on a satellite then the satellite will fall on the earth surface following a spiral path .

Escape velocity from moon = 2.4 km/sec where $g_m = \dfrac{g_c}{6}=1.6 m/s^2$.

An astronaut in a satellite cannot use a pendulum clock. However, he can use a spring clock or digital clock.

If two same material spheres of the same mass and radius (R) are kept in contact then the gravitational force between them is $F\propto R^4$

[$F = \dfrac{GM^2}{(2R)^2}= \dfrac{G (\dfrac{4}{3} \pi r^3 \rho )^2}{(2R)^2}$

$F \propto R^4$

Astronauts in space see the sky black due to the lack of atmosphere above them .

A bus weighs more while moving due west with sufficient speed than when it is at rest or is moving due east.

A satellite will be seen Geostationary only when it is launched at a proper height from West to East ( not east to west ) in an equatorial plane since the earth rotates itself west to east.

Don’t confuse that mass of a body at earths center is finite and positive but its weight is zero ( .. g= 0)

If a body falls freely infinite height it will reach the earth surface with thye escape velocity i.e 11.2 km/sec .

Greater is the height of satellite smaller is the orbital velocity. [ HINT : v_0= \sqrt{\dfrac{GM}{(R+h)} ]

Gravity holds the atmosphere around the earth.

A Satellite in vaccum doesn’t require any energy for orbiting.

Rocket launched with escape velocity follows a parabolic path.

The time period of an earth satellite in a circular orbit is independent of the radius of the satellite .

If a simple pendulum is taken from the equator to the pole its time period decreases. [Hint: $T \propto \dfrac{1}{\sqrt{g}} ; from the equator to pole value of g increases. ]

A person will get more quantity of matter in kg-wt at satellite where weight is minimum .

The acceleration due to gravity on the surface of earth varies directly with latitude .

In a gravitational field , the work done in transporting mass from one point to another depends on end positions since gravitational force is a conservative force ; work done depends upon the initial and final positions of the body and not on the body and not on the path traversed .

If the mass of planet is reduced , then orbital velocity of satellite decreases .

Mercury does not contain atmosphere . It is because of very low gravity and low temperature .

Moon has no atmosphere because r.m.s velocity of gas molecules there are greater than escape velocity of moon.

Gravity meter and Eotvos gravity balance are used to measure changes in acceleration due to gravity .

The line joining the places on earth having same values of g are called isogams .

Gravitational mass is proportional to force while internal mass to quantity of matter present in the body .

The minimum number of artificial satellites needed to be placed to the surface of earth for worldwide communication between any two location is 3 .

Time taken by radio wave to go and return back from communication satellite to earth is nearly ¼ sec .

Tidal waves in sea are principally due to gravitational effect of moon on earth .

If the weight of an object in a coal mine , at sea-level and at top of mountain are $W_1, W_2,$ and $W_3$ respectively , then $W_1 < W_2 > W_3$. It is because of g is maximum on the surface and decreases with depth or height .

Tail of comets point away from the sun because of the radiation pressure of the sun .

When the total energy of a satellite is –ve, it will be moving in either a circular or an elliptical orbit .

When then total energy of a satellite is zero, it will escape away from its orbit and the path becomes parabolic .

When a body falls freely from infinite height, it will be reaching the surface of the earth with a velocity of 11.2 kms$^{-1}$.

Orbital velocity of satellite at altitude n times the earth’s radius (i.e h =nR ) is $\dfrac{1}{\sqrt{(n+1)}}$ times the orbital velocity near earth’s surface .

Orbital velocity of satellite at orbit n times the earth’s radius ( i.e . $r = nR$) is $\dfrac{1}{\sqrt{n}}$ times near the near the earth’s surface .

Ratio of (Inertial mass)/(Gravitational mass) = 1

If mass of plant is constant and radius decreases by n% then g on surface decreases by 2n% . [ HINT : ∆g/g= -2∆R/R ]

If radius of planet is constant and mass increases by n% then g increases by n% . [Hint : ∆g/g = ∆M/M for R=constant

If radius of planet is constant and density of planet decreases by n% , the g on its surface [Hint : ∆g/g= ∆ρ/ρ ]

If the density of the planet is constant and radius decreases by n% then g decreases by n%.