Question

find the force of attraction between two spherical bodies of mass 10^ 7 kg each and having centre to centre distance of 10^ 7metre​

Answer 1

Given

• Mass₁ = Mass₂ = 10⁷ kg
• Distance = 10⁷ m

To Find

• Gravitational Force

Solution

☯ F = GMm/d²

• Here G is the gravitational constant whose value is 6.67 × 10⁻¹¹

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✭ According to the Question :

→ F = GMm/d²

→ F = (6.67 × 10⁻¹¹ × 10⁷ × 10⁷)/(10⁷)²

→ F = (6.67 × 10⁻¹¹ × 10¹⁴)/10¹⁴

→ F = (6.67 × 10⁽⁻¹¹ ⁺ ¹⁴⁾)/10¹⁴

→ F = 6.67 × 10³/10¹⁴

→ F = 6.67 × 10³ × 10⁻¹⁴

→ F = 6.67 × 10⁽³-¹⁴⁾

→ F = 6.67 × 10⁻¹¹ N

∴ The gravitational force in between is 6.67 × 10⁻¹¹ N

Answer 2

Given:-

➤The force of attraction between two spherical bodies of mass 10⁷ kg each.

➤Having a centre to centre distance of 10⁷ m.

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To Find:-

➤Force of attraction between two spherical bodies?...

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Solution:-

➤To find the force of attraction between two spherical bodies Using Formula,

$\longrightarrow{ \underline{ \boxed{ \tt{ \pink{F = \frac{GM}{ {d}^{2} } }}}}} \bigstar$

$\\ \longrightarrow \tt \: F = \frac{6.67 \times {10}^{ - 11} \times {10}^{7} \times {10}^{7} }{ {10}^{7} \times {10}^{7} }$

$\\ \longrightarrow \tt \: F = \frac{6.67 \times {10}^{ - 11} \times {10}^{7} \times {10}^{7} }{ ({10}^{7} ) {}^{2} }$

$\\ \longrightarrow \tt \: F = \frac{6.67 \times {10}^{ (- 11 + 14)} }{ {10}^{14}}$

$\\ \longrightarrow \tt \: F = \frac{6.67 \times {10}^{ 3} }{ {10}^{14}}$

$\\ \longrightarrow \tt \: F = {6.67 \times 10}^{ 3} \times {10}^{ - 14}$

$\\ \longrightarrow \tt \: F = {6.67 \times 10}^{3 - 14}$

$\\ \longrightarrow \tt \: F = {6.67 \times 10}^{- 11}$

➤The force of attraction between two spherical body:

$\longrightarrow{ \underline { \boxed {\tt \:{ F = {6.67 \times 10}^{- 11}}}}}$

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