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?

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Answer 1

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

✭ 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

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

Their ratios will be,

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

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

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

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

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

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

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

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Answer 2

Given:-

• Masses of two satellites are in the Ratio = 3:1
• Radius of their circular orbits are in the ratio = 1:4

Find:-

• Total Mechanical Energy of A and B

Solution:-

Let us suppose two satellites A and B having mass $\bf {m_1}$ and $\bf {m_2}$ and Radius of their circular orbits $\bf {r_1}$ and $\bf {r_2}$ respectively.

$\bigstar$ we, know that total mechanical energy of satellite is given by,

$\underline{\boxed{\sf E = - \: \dfrac{GMm}{2r}}}$

Now,

$\sf \to\dfrac{E_A}{E_B} = \dfrac{m_1}{r_1} \times \dfrac{r_2}{m_2}$

$\sf \to\dfrac{E_A}{E_B} = \dfrac{m_1}{ m_2} \times \dfrac{r_2}{r_1}$

where,

• Ratio of mass = 3:1
• Ratio of radii = 1:4

So,

$\sf \dashrightarrow\dfrac{E_A}{E_B} = \dfrac{m_1}{ m_2} \times \dfrac{r_2}{r_1}$

$\sf \dashrightarrow\dfrac{E_A}{E_B} = \dfrac{3}{1} \times \dfrac{4}{1}$

$\sf \dashrightarrow\dfrac{E_A}{E_B} = \dfrac{12}{1}$

$\small{ \sf \therefore \underline{E_A:E_B =12:1}}$

Hence, Ratio of Total Mechanical Energy of A and B is 12:1