Question

Earth Sun distance is held constant and the mass of sun is doubled then period of revolution of earth around sun

Answer 1

Dear Student,

◆ Answer -

T' = T/√2

● Explanation -

Period of revolution of earth around sun is given by -

T = 2π√(r^3/GM)

T ∝ 1/√M

T = k/√M

When mass of sun is doubled,

T' = k/√M'

T' = k/√(2M)

T' = (k/√M)/√2

T' = T/√2

Hence, period of revolution of earth around sun will reduce by √2 times.

Thanks dear. Hope this helps you...

Answer 2

$T' = \frac{T}{\sqrt2}$ is the period of revolution of earth around the sun.

Explanation:

The Size and mass of the sun decides the the the revolution time of the planet surrounding it.

Hence, when the size of the sun is doubled the gravitational pull of the Sun will also get doubled and this increased gravitational pull might take in some of the planets nearer to it or the planets shall fall in.

Period of revolution of earth around sun is given by,

T =$2\pi \sqrt{(\frac{r^3}{GM})}$

T ∝ $\frac{1}{\sqrt M}$

T = $\frac{k}{\sqrt M}$

When mass of sun is doubled,

T' = $\frac{k}{\sqrt{M'}}$

T' = $\frac{k}{\sqrt{2M}}$

T' = $\frac {\frac{k}{\sqrt M}}{\sqrt2}$

T' = $\frac{T}{\sqrt2}$

Then, the period of the earth's revolution around the sun will be reduced by $\sqrt2$ times.

To know more:

What will the duration of one year if distance between sun and Earth becomes double the present distance the present duration of year is 365 days.

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