Question

i.e the maximum portion
Nitrogen forms 60 y.
of the atmosphere​

Answer 1

Answer:

$\displaystyle\large\underline{\sf\red{Given}}$

✭ Mass of two Satellites are of the ratio 3:1

✭ Radii of their circular orbit are in the ratio 1:4

$\displaystyle\large\underline{\sf\blue{To \ Find}}$

◈ Ratio of their total mechanical energy?

$\displaystyle\large\underline{\sf\gray{Solution}}$

So here to find the total energy we may use,

$\displaystyle\sf \underline{\boxed{\sf Total \ Energy = \dfrac{-GMm}{2r}}}$

Also let the two Bodies be A & B

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$\underline{\bigstar\:\textsf{According to the given Question :}}$

We are given that,

⪼ $\displaystyle\sf \dfrac{m_1}{m_2} = \dfrac{3}{1}$

And,

⪼ $\displaystyle\sf \dfrac{r_1}{r_2} = \dfrac{1}{4}$

So then their total energy (E) will be,

➝ $\displaystyle\sf E_A = \dfrac{-GMm_1}{2r_1}$

And

➝$\displaystyle\sf E_B = \dfrac{-GMm_2}{2r_2}$

➳$\displaystyle\sf \dfrac{\dfrac{-GMm_1}{2r_1}}{\dfrac{-GMm_2}{2r_2}}$

➳$\displaystyle\sf \dfrac{m_1}{r_1} \times \dfrac{r_2}{m_2}$

➳ $\displaystyle\sf \dfrac{m_1}{m_2} \times \dfrac{r_2}{r_1}$

➳ $\displaystyle\sf \dfrac{3}{1}\times \dfrac{4}{1}$

➳ $\displaystyle\sf \dfrac{3\times4}{1}$

➳$\displaystyle\sf\pink{\dfrac{E_A}{E_B} = \dfrac{12}{1}}$

$\displaystyle\sf \therefore\:\underline{\sf Their \ Ratio \ will \ be \ E_A:E_B = 12:1}$

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