Question

The mass of the Sun is 2 x 10³⁰ kg and that of the Earth is 6 x10²⁴ kg. If the average distance between the Sun and the Earth
is 1.5 x 10¹¹ m, calculate the force exerted by the Sun on the
Earth and also by Earth on the Sun.

Answer 1

$\star \; {\underline{\boxed{\pmb{\orange{\sf{ \; Given \; :- }}}}}}$

• Mass of Sun = $\sf{ 2 \times 10^{30} \; kg }$
• Mass of Earth = $\sf{ 6 \times 10^{24} \; kg }$
• Distance = $\sf{ 1.5 \times 10^{11} \; m }$

$\\ \qquad{\rule{200pt}{2pt}}$

$\star \; {\underline{\boxed{\pmb{\green{\sf{ \; To \; Find \; :- }}}}}}$

• Force Exerted = ?

$\\ \qquad{\rule{200pt}{2pt}}$

$\star \; {\underline{\boxed{\pmb{\pink{\sf{ \; SolutioN \; :- }}}}}}$

$\maltese \; {\underline{\pmb{\textbf{\textsf{ Formula \; Used \; :- }}}}}$

• ${\underline{\boxed{\pmb{\sf{ F = G \dfrac{ M_1 \times M_2 }{ r^{2} } }}}}}$

Where :

• F = Force Exerted
• G = Universal Gravitational Constant
• M¹ = Mass of Earth
• M² = Mass of Sun
• r = Distance between them

$\maltese \; {\underline{\pmb{\textbf{\textsf{ \; Calculating the Force Exerted :- }}}}}$

$\; {\dashrightarrow{\qquad{\sf{ F = G \dfrac{ M_1 \times M_2 }{ r^{2} } }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ F = \dfrac{ \bigg( 6.67 \times {10}^{-11} \bigg) \times \bigg( 6 \times {10}^{24} \bigg) \times \bigg( 2 \times {10}^{30} \bigg) }{ \bigg( 1.5 \times {10}^{11} \bigg)^{2} } }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ F = \dfrac{ 80.04 \times {10}^{43} }{ 2.25 \times {10}^{22} } }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ F = \dfrac{ 80.04 \times {10}^{43} }{ 2.25 \times {10}^{22} } }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ F = 35.57 \times 10^{(43 - 22)} }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ F = 35.57 \times 10^{21} }}}} \\ \\ \\ \ {\qquad \; \; {\therefore \; {\underline{\boxed{\pmb{\red{\sf{ 3.57 \times {10}^{22} \; N }}}}}}}} \; {\purple{\bigstar}}$

$\therefore \; \;$ Force Exerted by sun on Earth is 3.57 × 10²² Newton .

$\\ \qquad{\rule{200pt}{2pt}}$

Answer 2

$\large \dag \: \frak {\underline{ \pink{Concept :-}}}$

⠀

To solve this question we will use the concept of "Gravitation", from class 9th as per NCERT. More specifically, we will use the concept of "Universal Law of gravitation". This law states that, force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of distance between them. This givens us the formula :-

⠀

$\boxed{ \sf{F_{attraction} \propto\dfrac{M_1 \times M_2}{ {R}^{2} } }}$

⠀

Now, this sign of proportionality is changed into sign of equality, by using a constant, known as gravitational constant, that is "G", it has a specific value of $\underline{ \boxed{\sf 6.67 x 10 ^{ - 11} }}$. So, the formula becomes :-

⠀

$\qquad \star\: \blue{\underline{ \boxed{ \sf F_{attraction} = G\dfrac {M_1 \times M_2} {R ^{2} }}} }\: \star$

⠀

Where,

⠀

• G = Universal Gravitational Constant
• M1 = Mass of Object 1
• M2 = Mass of object 2
• R = Distance between those

⠀

Now we will start with the question.

⠀

────────────────────────

⠀

$\large \dag \: \underline \red{\mathfrak{Given :-}}$

⠀

• Mass of the Sun = 2 x 10³⁰ kg
• Mass of the Earth = 6 x10²⁴
• Distance between them = 1.5 x 10¹¹ m

⠀

$\large \dag \: \underline{\purple{ \mathfrak{To ~Find :-}}}$

⠀

• The Force of attraction between them

⠀

$\large \dag \: \underline{ \orange{ \mathfrak{Elucidation :-}}}$

⠀

Putting the values we get :-

⠀

$\\ \\ \sf : \dashrightarrow \: F_{attraction} = \dfrac{6.67 \times {10}^{ - 11} \times 2 \times {10}^{30} \times 6 \times {10}^{24} }{ \{1.5 \times {10}^{11} \} ^{2} } \\ \\ \\ \sf : \dashrightarrow \: F_{attraction} = \dfrac{6.67 \times 2 \times 6 \times {10}^{ \{ - 11 + 30 + 24 \}} }{ 2.25 \times {10}^{22} } \\ \\ \\ \sf : \dashrightarrow \: F_{attraction} = \dfrac{6.67 \times 2 \times 6 \times {10}^{ 43 } }{ 2.25 \times {10}^{22} } \\ \\ \\ \sf : \dashrightarrow \: F_{attraction} = \dfrac{6.67 \times 2 \times 6 \times {10}^{ 43 - 22} }{ 2.25 } \\ \\ \\ \sf : \dashrightarrow \: F_{attraction} = \dfrac{6.67 \times 2 \times 6 \times {10}^{ 21} }{ 2.25 } \\ \\ \\ \sf : \dashrightarrow \: F_{attraction} = \star\:\boxed{\sf 3.57 \times {10}^{22}} ~\star$

⠀

──────────────────

⠀