Question

Two satellites have their masses in the ratio 3:1. The radii of their circular orbits are in the ratio 1:4. What is the total mechanical energy of A and B?


Correct answer with proper explanation...​

Answer 1

Answer:

hope it helps

Step-by-step explanation:

Thanks for the question

Answer 2

Given,

» Ratio of radius of orbits of satellite A and B = 3 : 1

» Ratio of masses of satellite A and B = 1 : 4

• Let x be the constant by which both the radius of orbits of satellites should be multiplied to attain original radius of satellite A and B.

• Let y be the constant by which both the masses should be multiplied to attain original masses of satellite A and B.

Radius of orbit of satellite A = 3x

Mass of satellite A = 1x

Radius of orbit of satellite B = 1x

Mass of satellite B = 4x

» Mass of the Earth = M (constant for both satellites)

» Gravitational Constant = G

» Total Mechanical Energy = E

» Mass of the satellite = m

» Radius of orbit of satellite = r

______________________________________________

~ For Total Mechanical Energy of satellite A and B ::

We know that,

$\\\;\sf{:\rightarrow\;\;E\;=\;\bf{-\:\dfrac{G\:M\:m}{2r}}}$

By applying values,

• For Satellite A ::

$\\\;\sf{:\Longrightarrow\;\;E_{A}\;=\;\bf{-\:\dfrac{G\:M\:3x}{2(1y)}}}$

$\\\;\sf{:\Longrightarrow\;\;\blue{E_{A}\;=\;\bf{-\:\dfrac{G\:M\:3x}{2y}}}}$

• For Satellite B ::

$\\\;\sf{:\Longrightarrow\;\;E_{B}\;=\;\bf{-\:\dfrac{G\:M\:1x}{2(4y)}}}$

$\\\;\sf{:\Longrightarrow\;\;\green{E_{B}\;=\;\bf{-\:\dfrac{G\:M\:x}{8y}}}}$

______________________________________________

~ For ratio of total mechanical energy of two satellites ::

This is given as,

$\\\;\sf{\mapsto\;\;\red{Ratio\;of\;Mechanical\;Energies\;=\;\bf{\dfrac{E_{A}}{E_{B}}}}}$

By applying values, we get

$\\\;\sf{\mapsto\;\;\dfrac{E_{A}}{E_{B}}\;=\;\bf{\dfrac{\bigg(-\:\dfrac{G\:M\:\times\:3x}{2y}\bigg)}{\bigg(-\:\dfrac{G\:M\:\times\:x}{8y}\bigg)}}}$

Cancelling the -ve sign, we get

$\\\;\sf{\mapsto\;\;\dfrac{E_{A}}{E_{B}}\;=\;\bf{\dfrac{(G\:M\:\times\:3x\;\times\;8y)}{(G\:M\:\times\:x\;\times\;2y)}}}$

Cancelling G and M, we get

$\\\;\sf{\mapsto\;\;\dfrac{E_{A}}{E_{B}}\;=\;\bf{\dfrac{3x\;\times\;8y}{x\;\times\;2y}}}$

Cancelling x , we get

$\\\;\sf{\mapsto\;\;\dfrac{E_{A}}{E_{B}}\;=\;\bf{\dfrac{3\;\times\;8y}{1\;\times\;2y}}}$

Cancelling y, we get

$\\\;\sf{\mapsto\;\;\dfrac{E_{A}}{E_{B}}\;=\;\bf{\dfrac{3\;\times\;8}{2}}}$

$\\\;\sf{\mapsto\;\;\dfrac{E_{A}}{E_{B}}\;=\;\bf{\dfrac{3\;\times\;4}{1}}}$

$\\\;\bf{\mapsto\;\;\dfrac{E_{A}}{E_{B}}\;=\;\bf{\orange{\dfrac{12}{1}}}}$

In ratio form, this can be written as

$\\\;\bf{\mapsto\;\;E_{A}\;:\;{E_{B}\;=\;\bf{\purple{12\;:\;1}}}}$

$\\\;\underline{\boxed{\tt{Hence,\;\:ratio\;\:of\;\:Total\;\:mechanical\;\:Energy\;\:=\:\bf{\purple{12\;:\;1}}}}}$