Question

Q. The mass of the Earth is 6 × 10²⁴ kg. The distance between the Earth and Sun is 1.5 × 10¹¹ m. If the Gravitational Force between the two is 3.5 × 10²² N, What is mass of the Sun?

$\huge{\underline{\sf{\green{Key \: Points-}}}}$

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$\boxed{\sf \red{★ \: Use \: G \: = \: 6.7 \times {10}^{ - 11n {m}^{2} {kg}^{2} }}}$

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$\boxed {\ \sf \blue{Ans:1.96 \times {10}^{30} kg}}$


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Answer 1

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Answer 2

Mass of the sun = 1.967 × 10³⁰ kg

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$\orange{\bigstar}★ Given \green{\bigstar}★$

The mass of the Earth is 6 × 10²⁴ kg

The distance between the Earth and Sun is 1.5 × 10¹¹ m

The Gravitational Force between the two is 3.5 × 10²² N

$\orange{\bigstar}★ To Find \green{\bigstar}★$

Mass of the Sun , M

$\orange{\bigstar}★ Knowledge required \green{\bigstar}★$

The gravitational force between two objects is given by ,

$\bf \pink{\bigstar}\ \; F=\dfrac{GMm}{r^2}★$

where ,

F denotes gravitational force

G denotes gravitational constant

M denotes mass of the sun

m denotes mass of the earth

r denotes distance between M and m

$\orange{\bigstar}★ Solution \green{\bigstar}★$

Given ,

m = 6 × 10²⁴ kg

M = ? kg

r = 1.5 × 10¹¹ m

G = 6.7 × 10⁻¹¹ Nm²kg²

F = 3.5 × 10²² N

Apply formula ,

$\begin{gathered}\bf \orange{\red{\bigstar}\ \;F=\dfrac{GMm}{r^2}}\\\\\to\ \rm 3.5\times 10^{22}=\dfrac{6.67\times 10^{-11}\times M\times 6\times 10^{24}}{(1.5\times 10^{11})^2}\end{gathered}$

$\begin{gathered}\to\ \rm 3.5\times 10^{22}=\dfrac{40.02\times 10^{13}\times M}{2.25\times 10^{22}}\\\\\to \rm 3.5\times 10^{22}=17.7867\times 10^{-9}\times M\\\\\to \rm M=\dfrac{3.5\times 10^{22}}{17.7867\times 10^{-9}}\\\\\to \rm M=0.1967\times 10^{31}\\\\\to \bf M=1.967\times 10^{30}\ kg\ \; \pink{\bigstar}\end{gathered}$

So , Mass of the sun = 1.967 × 10³⁰ kg

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