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 will be appreciated,spams will be reported...​

Answer 1

Answer:

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?

Explanation:

hope it helps you. please make me brainlest and thank me

Answer 2

★ Correct 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 ratio of total mechanical energy of A and B?

______________________________________________

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the Concept of Total Mechanical Energy has been used. We see that here in the question its given that there are two satellites, this means we can use the formula of Total Mechanical Energy which is the sum of Gravitational Potential energy and Gravitational Kinetic Energy. Firstly we can apply the values we have in the equation and then find its ratio to get out answer.

Let's do it !!

______________________________________________

★ Formula Used :-

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

______________________________________________

★ Solution :-

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

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

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

______________________________________________

★ More to know :-

$\\\;\sf{\leadsto\;\;F\;=\;\dfrac{Gm\:m_{r}}{r^{2}}}$

$\\\;\sf{\leadsto\;\;Gravitational\;Potential\;Energy\;=\;\dfrac{-\:GMm}{r}}$

$\\\;\sf{\leadsto\;\;Gravitational\;Potential\;=\;\dfrac{-\:GM}{r}}$

$\\\;\sf{\leadsto\;\;Intensity\;of\;Gravitational\;Field\;=\;\dfrac{GM}{r^{2}}}$

$\\\;\sf{\leadsto\;\;E\;=\;\dfrac{F}{m}}$