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>> $Newton's Law of Gravitation from Kepler'...$

Newton's Law of Gravitation from Kepler's Law of Periods

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: Kepler's third law states that square of period of revolution \$(T)\$ of a planet around the sun, is proportional to third power of average distance \$r\$ between sun and planeti.e \$T^2=Kr^3\$ here \$K\$ is constant.If the masses of sun and planet are \$M\$ and \$m\$ respectively than as per Newton's law of gravitation force of attraction between them is\$F=\dfrac {GMm}{r^2}\$, here \$G\$ is gravitational constant The relation between \$G\$ and \$K\$ is described as

The gravitational force is providing centripetal force to the planets oribital circular motion\$\dfrac{GMm}{r^2}=\dfrac{mv^2}{r}\$\$\Rightarrow v=\sqrt{\dfrac{GM}{r}}\$The time period for planet in one revolution will be \$T=\dfrac{2\pi r}{v}=\dfrac{2\pi r}{\sqrt{\dfrac{GM}{r}}}\$squaring each side\$T^2=\dfrac{4\pi^2 r^3}{GM}\$comparing with given equation \$T^2=Kr^3\$\$K=\dfrac{4\pi^2}{GM}\$\$KGM=4\pi^2\$hence, correct answer is option D

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