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This is second of our many quizzes on Universal gravitation. Click on this to access more such competitive quizzes on Mechanics.

Universal gravitation Quiz 2

physics

If the density $\rho $ of two planet is same than the ratio of acceleration due to gravity on the surface of planet 1 of radius ${{R}_{A}}$ to the planet 2 of radius ${{R}_{B}}$ will be- 

 If the density of planets is equal then their acceleration due to gravity will depend on the radius of the planet. 

  $g=\dfrac{GM}{{{R}^{2}}}\\  \Rightarrow {{\rho }_{A}}=\dfrac{M}{4\pi {{R}_{A}}^{3}}\\  \Rightarrow {{\rho }_{B}}=\dfrac{M}{4\pi {{R}_{B}}^{3}}\\  \Rightarrow {{\rho }_{B}}={{\rho }_{A}}\\  \Rightarrow\dfrac{M}{4\pi {{R}_{B}}^{3}}=\dfrac{M}{4\pi {{R}_{A}}^{3}}\\  \text{Multiply by G both side we have-}\\  \dfrac{GM}{4\pi {{R}_{B}}^{3}}=\dfrac{GM}{4\pi {{R}_{A}}^{3}}\\  \Rightarrow \dfrac{{{g}_{B}}}{{{R}_{B}}}=\dfrac{{{g}_{A}}}{{{R}_{A}}}\\ \therefore \dfrac{{{g}_{A}}}{{{g}_{B}}}=\dfrac{{{R}_{A}}}{{{R}_{B}}}$

 If a planet is revolving around a sun in a orbit of radius $r$ and time period $T$ then the correct relationship between $T$ and $r$ is given by- 

The proportionality is power of $\dfrac{3}{2}$ so the graph will not be straight line instead of that it will be curved and increasing. 

  ${{T}^{2}}\propto {{R}^{3}}\\  \therefore T\propto {{R}^{3/2}}\\  \text{The graph will be curved and increasing.}$

 The minimum velocity by which an object must be thrown vertically upward so that it can reach up to a height of $8R$, here $R$ is the radius of earth equal to $6.4\times {{10}^{8}}m$.$[\text{Mass of earth}\,M=6\times {{10}^{24}}\,kg\,\text{and Gravitational constant}\,G=6.7\times {{10}^{-11}}N-{{m}^{2}}/k{{g}^{2}}]$ 

The kinetic energy and potential energy at the surface of earth is used in conservation of energy as it will be converted into potential energy only. 

 $  \text{Applying energy conservation}\\  \dfrac{1}{2}m{{v}^{2}}-\dfrac{GMm}{R}=-\dfrac{GMm}{9R}\\  \Rightarrow \dfrac{1}{2}{{v}^{2}}=\dfrac{GM}{R}-\dfrac{GM}{9R}\\  \Rightarrow \dfrac{1}{2}{{v}^{2}}=\dfrac{8GM}{9R}\\  \Rightarrow \dfrac{1}{2}{{v}^{2}}=\dfrac{8\times 6.7\times {{10}^{-11}}\times 6\times {{10}^{24}}}{9\times 6.4\times {{10}^{8}}}\\  \Rightarrow \dfrac{1}{2}{{v}^{2}}=5.58\times {{10}^{5}}\\  \Rightarrow {{v}^{2}}=1.11\times {{10}^{6}}\\  \therefore v=10.56\times {{10}^{3}}\,m/s$

If an object is taken down to depth $d=R/4$ from the surface of earth. The percentage change in acceleration due to gravity will be $[\text{assuming acceleration due to gravity on the surface is g and radius of earth is R}]$<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/8b3d9e13-fd3d-4343-b462-669edc0dd03d2275005666944677990.png"/> 

 Acceleration due to gravity will decrease while moving away from the surface or going below the surface. 

 $g=\dfrac{GM}{{{r}^{2}}}\\  \text{Gravity at distance}\,\dfrac{R}{4}\,\text{below the surface will be-}\\ g'=g\left( 1-\dfrac{d}{R} \right)\\  \Rightarrow g' =g\left( 1-\dfrac{R/4}{R} \right)\\  \Rightarrow g' =\dfrac{3g}{4}\\  \text{Percentage decrease}=\dfrac{g-3g/4}{g}\times 100\\  \therefore\text{Percentage decrease}=25\%$

An object is placed between two planets of mass $M$ and $4M$ at a distance of $R$ and $4R$ from the planets. The radius of the smaller planet is $R$ and that of the larger planet is $2R$. Find the effective gravitational field experienced by object- 

Both planets will try to pull objects with their force of gravity so that the field on the object will be in the opposite direction. 

 $ E=\dfrac{GMm}{{{x}^{2}}}\\  \text{Field experienced by object-}\\ E=\dfrac{GM}{{{(R+R)}^{2}}}-\dfrac{4GM}{{{(4R+2R)}^{2}}}\\  \Rightarrow E =\dfrac{GM}{4{{R}^{2}}}-\dfrac{4GM}{36{{R}^{2}}}\\  \therefore E =\dfrac{5GM}{36{{R}^{2}}}$