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Universal Law of Gravitation - L1

Q: Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth\$=6\times 10^{24}\$kg and of the Sun\$=2\times 10^{30}kg\$. The average distance between the two is \$1.5\times 10^{11}\$m.

Given : Mass of earth \$m = 6\times 10^{24} kg\$ Mass of Sun \$M = 2\times 10^{30} kg\$ Distance between sun and earth \$r = 1.5 \times 10^{11}\$ mForce of gravitation between them \$F = \dfrac{GMm}{r^2}\$\$F = \dfrac{6.67 \times 10^{-11} \times (6\times 10^{24}) \times (2\times 10^{30})}{(1.5 \times 10^{11})^2} = 3.557 \times 10^{22}\$ N

Q: Two particles of equal mass m go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

Given,Masses of the two particles are \$m_1=m_2=m\$And radius of the two particles are \$r_1=r_2=R\$So ,The Gravitational force of attraction between the particles \$F=\dfrac{G\times{m}\times{m}}{R^2}--------(1)\$Ans, centripital force \$F=\dfrac{m\times{v^2}}{R}------(2)\$From given condition (1) and (2)\$\dfrac{Gmm}{(2R)^2}=\dfrac{mv^2}{R}\$, \$v=\dfrac{Gm}{4R}=\dfrac{1}{2}\sqrt{\dfrac{Gm}{R}}\$So the speed of the each particle is \$v=\dfrac{1}{2}\sqrt{\dfrac{Gm}{R}}\$

Q: Two spherical bodies of mass M and 5M and radii R and 2R respectively are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only. then the distance covered by the smaller body just before collision is

Distance between their surfaces \$= 12R -R -2R = 9R\$Since, P \$\propto\$ massa \$\propto\$ massDistance \$\propto\$ accelerationSo, we can write;\$\cfrac{a_1}{a_2} = \cfrac{m}{5m} = \cfrac{s_1}{s_2}\$\$ \cfrac{s_1}{s_2} = \cfrac {1}{5}\$\$5s_1 = s_2\$ .......................(1)\$s_1 + s_2 = 9R \$ .......................(2)On solving these equations;\$s_1 = 1.5R\$\$s_2 = 7.5R\$Since smaller ball have more acceleration in same time interval, smaller ball will cover more distance.

Q: If the distance between centres of the earth and the moon is \$d\$ and the mass of the earth is \$81\$ times the mass of the moon, then at what distance from centre of the earth, the gravitational field will be zero?

Lets assume there is a particle of mass \$m\$ from \$x\$ distance away from earth. assuming gravitational force on \$m\$ is zero so :-.Force balance\$ \dfrac{(Gm81M)}{x^2} \$ = \$ \dfrac{(G m M)}{(d-x)^2} \$\$\dfrac{81}{x^2} \$ = \$ \dfrac{1}{(d-x)^2} \$\$\dfrac{x}{9} \$ = \$ \dfrac{d-x}{1} \$solving above wquation we will get\$x = \dfrac{9d}{10} \$

Q: What is the magnitude of the gravitational force between the earth and a \$1\ kg\$ object on its surface? (Mass of the earth is \$6\times 10^{24}\ kg\$ and radius of the earth is \$6.4\times 10^6\ m\$.)

Given : Mass of earth \$M = 6\times 10^{24}\ kg\$ Mass of object \$m = 1\ kg\$ Radius of earth \$R = 6.4 \times 10^{6}\ m\$ Force of gravitation between them, \$F = \dfrac{GMm}{R^2}\$\$F = \dfrac{6.67 \times 10^{-11} \times (6\times 10^{24}) \times 1}{(6.4 \times 10^{6})^2} = 9.77\ N\$

Q: Two spheres of masses m and M are situated in air and the gravitational force between them is F. The space around the masses is now filled with a liquid of specific gravity \$3\$. The new gravitational force will be :

The gravitational force will remain the same as it is independent of density of medium.

Q: Dimensional formula of universal gravitational constant \$G\$ is-

Gravitational force acting between two objects of masses \$m_1\$ and \$m_2\$ separated by distance \$r\$, \$F = \dfrac{Gm_1 m_2 }{r^2}\$\$\implies\$ \$G = \dfrac{Fr^2}{m_1 m_2}\$Thus dimensional formula of \$G\$ is \$\dfrac{[M L T^{-2}] [L]^2}{[M]^2}\$\$\implies\$ \$G = M^{-1} L^3 T^{-2}\$

Q: A geostationary satellite is orbiting the earth at a height of \$5R\$ above the surface of the earth has a time period of \$24\$ hours. \$R\$ being the radius of the earth.The time period of another satellite in hours at a height \$2R\$ from the surface of the earth is

\$\textbf{Step 1: Kepler's Third Law} \$From Kepler's Third Law "The square of the time period \$(T)\$ of revolution of a planet around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis \$(r)\$" \$T^2 \propto r^3\$So, \$T_{1}^2 \propto r_{1}^3\$ and \$T_{2}^2 \propto r_{2}^3\$ \$\therefore \ \ T_2 = T_1 \left(\dfrac{r_2}{r_1}\right)^{\frac{3}{2}}\$ \$....(1)\$\$\textbf{Step 2: Time period Calculation}\$Given, Time period of first satellite \$T_1 = 24 hr\$Orbiting radius of first satellite \$r_1 = 5R + R = 6R\$Orbiting radius of second satellite \$r_2 = 2R + R = 3R\$\$\therefore\$ From Equation \$(1)\$, Time period of second satellite: \$T_2 = 24\ hr \left(\dfrac{3R}{6R}\right)^{\frac{3}{2}} = \dfrac{24}{2\sqrt{2}} hours\$ \$T_2 = 6\sqrt{2}\ hours\$Hence time period of another satellite in hours at height \$2R\$ from the surface of earth is \$6\sqrt{2}\ hours\$. Option \$(A)\$

Q: A mass \$M\$ is split into two parts \$m\$ and \$\left( M-m \right) \$, which are, then separated by a certain distance. The ratio \${ m }/{ M }\$ which maximizes the gravitational force between the parts is

Gravitational force between two parts \$F = \dfrac{Gm(M-m)}{r^2}\$For maxima, \$\dfrac{dF}{dm} =0\$Or \$ \dfrac{d}{dm} .\dfrac{Gm(M-m)}{r^2} = 0\$Or \$ \dfrac{d}{dm}(mM-m^2)= 0\$Or \$M-2m = 0\$ \$\implies m = \dfrac{M}{2}\$Hence force between them is maximum when \$m:M = 1:2\$

Q: Imagine a light planet revolving around a very massive star in a circular orbit of radius r with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to \$r^{5/2}\$, then the square of the time period will be proportional to.

Gravitational force,\${ F }_{ G }\Rightarrow \cfrac { 1 }{ { r }^{ 5/2 } } \$revolvong in circular orbit, so, \${ F }_{ G }=\$centripetal forceCentripetal force\$\propto { r }^{ -5/2 }\$\$mr{ \omega }^{ 2 }\propto { r }^{ -5/2 }\$Mass is constant, so,\${ \omega }^{ 2 }\propto \cfrac { { r }^{ -5/2 } }{ r } \$\${ \omega }^{ 2 }\propto { r }^{ -7/2 }\quad \omega =\cfrac { 2\pi }{ T } \$\${ \left( \cfrac { 2\pi }{ T } \right) }^{ 2 }\propto { r }^{ -7/2 }\$\${ T }^{ 2 }\propto { r }^{ 7/2 }\$

Physics

example

Problems involving universal law of gravitation

Example: If

F_{g}

and

F_{e}

are gravitational and electrostatic forces between two electrons at a distance

0.1

m

then

F_{g} / F_{e}

is in what order?

Solution: Gravitational force

F_{g} = \frac{G m ^{2}}{r ^{2}}

Electrostatic force

F_{e} = \frac{K q ^{2}}{r ^{2}}

\frac{F _{g}}{F _{e}} = \frac{G m ^{2}}{K q ^{2}}

where

G =

Universal gravitational constant and

K = 1 / 4 π ϵ_{0}

G = 6.67 \times 1 0^{- 11}

and

K = 9 \times 1 0^{9}

Mass of an electron is

9.31 \times 1 0^{- 31}

and Charge of an electron is

1.6 \times 1 0^{- 19}

Hence,

F_{g} / F_{e} = G m^{2} / K q^{2}

is of the order of

1 0^{- 11 - 62 - 9 + 38} = 1 0^{- 43}

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Imagine a light planet revolving around a very massive star in a circular orbit of radius r with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to $r^{5 / 2}$ , then the square of the time period will be proportional to.

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Problems involving universal law of gravitation

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Universal Law of Gravitation - L1

Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth=6×1024kg and of the Sun=2×1030kg. The average distance between the two is 1.5×1011m.

Two particles of equal mass m go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

Two spherical bodies of mass M and 5M and radii R and 2R respectively are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only. then the distance covered by the smaller body just before collision is

If the distance between centres of the earth and the moon is d and the mass of the earth is 81 times the mass of the moon, then at what distance from centre of the earth, the gravitational field will be zero?

What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6×1024 kg and radius of the earth is 6.4×106 m.)

Two spheres of masses m and M are situated in air and the gravitational force between them is F. The space around the masses is now filled with a liquid of specific gravity 3. The new gravitational force will be :

Dimensional formula of universal gravitational constant G is-

A geostationary satellite is orbiting the earth at a height of 5R above the surface of the earth has a time period of 24 hours. R being the radius of the earth.The time period of another satellite in hours at a height 2R from the surface of the earth is

A mass M is split into two parts m and (M−m), which are, then separated by a certain distance. The ratio m/M which maximizes the gravitational force between the parts is

Imagine a light planet revolving around a very massive star in a circular orbit of radius r with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to r5/2, then the square of the time period will be proportional to.

Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth $= 6 \times 1 0^{24}$ kg and of the Sun $= 2 \times 1 0^{30} k g$ . The average distance between the two is $1.5 \times 1 0^{11}$ m.

If the distance between centres of the earth and the moon is $d$ and the mass of the earth is $81$ times the mass of the moon, then at what distance from centre of the earth, the gravitational field will be zero?

What is the magnitude of the gravitational force between the earth and a $1 k g$ object on its surface? (Mass of the earth is $6 \times 1 0^{24} k g$ and radius of the earth is $6.4 \times 1 0^{6} m$ .)

Two spheres of masses m and M are situated in air and the gravitational force between them is F. The space around the masses is now filled with a liquid of specific gravity $3$ . The new gravitational force will be :

Dimensional formula of universal gravitational constant $G$ is-

A geostationary satellite is orbiting the earth at a height of $5 R$ above the surface of the earth has a time period of $24$ hours. $R$ being the radius of the earth.The time period of another satellite in hours at a height $2 R$ from the surface of the earth is

A mass $M$ is split into two parts $m$ and $(M - m)$ , which are, then separated by a certain distance. The ratio $m / M$ which maximizes the gravitational force between the parts is

Imagine a light planet revolving around a very massive star in a circular orbit of radius r with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to $r^{5 / 2}$ , then the square of the time period will be proportional to.