Question

⚠️Don't spam⚠️ Calculate the force of Attraction due to the earth on a ball of 1 kg mass lying in the ground. [Mass of Earth = 6 × 10²⁴ kg , Radius of Earth = 6.4 × 10³ km, G = 6.7 × 10^-8 N m²/kg². It's a hard Question, but can be done Please help me

Answer 1

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${\red { \bf Question}}$

• Calculate the force of Attraction due to the earth on a ball of 1 kg mass lying in the ground. [Mass of Earth = 6 × 10²⁴ kg , Radius of Earth = 6.4 × 10³ km, G = 6.7 × 10⁻¹¹ N m²/kg²]

$\underline {\underline{\purple { \sf Given}}}$

• Mass of Ball (m₁) = 1 kg
• Mass of Earth (m₂)= 6 × 10²⁴ kg
• Radius of Earth (r)= 6.4 × 10³ km
• Gravitational Constant (G)= 6.7 × 10⁻⁸ N m²/kg².

$\underline {\underline{\purple { \sf To \: Find}}}$

• Force of attraction due to the earth

$\red{ \sf Mass \: of \: Ball = m _1}$

$\red{ \sf Mass \: of \: Earth = m _2}$

____________________

$\orange{ \sf m_1 = 1 \: Kg}$

$\orange{ \sf m_2= 6 \times 10^{24} \: km}$

➥ $\sf r = 6.4 \times 10^{3} \: km$

➥ $\sf r = 6.4 \times 10^{3} \times 1000 \: m$

➥ $\sf r = 6.4 \times 10^{6} \: m$

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☞ $\sf F = 6.7 \times {10}^{ - 11} \times \dfrac{1 \times 6 \times {10}^{24} }{(6.4 \times 6.4 \times {10}^{6})^{2} }$

☞ $\sf F = \dfrac{6.7 \times {10}^{ - 11} \times 6 \times {10}^{24} }{6.4 \times 6.4 \times {10}^{6} \times {10}^{6} }$

☞ $\sf F = \dfrac{6.7 \times 6 \times {10}^{ - 11} \times {10}^{24} }{6.4 \times 6.4 \times {10}^{6} \times {10}^{6} }$

☞ $\sf F = \dfrac{6.7 \times 6 \times {10}^{ - 13} }{6.4 \times 6.4 \times {10}^{ 6} \times {10}^{6} }$

☞ $\sf F = \dfrac{6.7 \times 6 \times {10}^{ 13} }{6.4 \times 6.4 \times {10}^{12} }$

☞ $\sf F = \dfrac{6.7 \times 6 \times {10} }{6.4 \times 6.4 }$

☞ $\sf F = \dfrac{67 \times 6 \times {10} \times 10 \times \cancel{10} }{64 \times 64 \times \cancel{10} }$

☞ $\sf F = \dfrac{67 \times 6 \times {10} \times 10 }{64 \times 64 }$

☞ $\sf F = \dfrac{40200 }{4096 }$

☞ $\boxed{\sf F = 9.8}$

Therefore, 9.8 Newton

$\underline { \underline{ \red { \sf Three : Equations \: of \: Motion}}}$

• $\blue{\sf v = u + at}$
• $\blue{\sf s = ut + \dfrac{1}{2} at ^{2}}$
• $\blue{\sf v^{2}-u^{2} = 2as}$

$\underline { \underline{ \green { \sf Some \: Important \: Formulas}}}$

• $\blue{ \sf F = \dfrac{Gm_1 \times Gm_2}{R^{2} }}$

• $\blue{\sf g = \dfrac{Gm}{R^{2} }}$

• $\blue{\sf v = u + gt}$

• $\blue {\sf v^{2} - u^{2} = 2gs}$

• $\blue {\sf s = ut + \dfrac{1}{2} gt^{2} }$

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Answer 2

Correct Question :

• Calculate the force of Attraction due to the earth on a ball of 1 kg mass lying in the ground. [Mass of Earth = 6 × 10²⁴ kg , Radius of Earth = 6.4 × 10³ km, G = 6.7 × 10⁻¹¹ N m²/kg².

[ The value of universal gravitational constant G had been found to be 6.7 × 10⁻¹¹ N m²/kg² ]

Given :

• Mass of the earth, M = 6 × 10²⁴ kg

• Mass of the ball, m = 1 kg

• Gravitational constant, G = 6.7 × 10⁻¹¹ Nm²/kg²

• Radius of the earth = 6.4 × 10³ km

To Find :

• Force of attraction due to the earth on the ball

Solution :

First of all, let's convert the given radius into it's standard form :

→ 6.4 × 10³ km

→ 6.4 × 10³ × 1000 m

→ 6.4 × 10⁶ m

Now,

To calculate the gravitational Force anywhere in this universe, the formula used is :

$\bigstar \: \sf \: F = G \: \times \dfrac{M \times m}{ {r}^{2} }$

$\sf \longrightarrow \: F = \dfrac{6.7 \times {10}^{ - 11} \times 6 \times {10}^{24} \times 1 }{ {(6.4 \times {10}^{6} })^{2} }$

$\sf \longrightarrow \: F = \dfrac{6.7 \times 6 \times 1 \times {10}^{ - 11} \times {10}^{24} }{6.4 \times 6.4 \times {10}^{6} \times {10}^{6} }$

$\sf \longrightarrow \: F = \dfrac{40.2 \times {10}^{ - 11 + 24} }{40.96 \times {10}^{6 + 6} }$

$\sf \longrightarrow \: F = \dfrac{40.2 \times {10}^{13} }{40.96 \times {10}^{12} }$

$\sf \longrightarrow \: F = \dfrac{40.2}{40.96} \times ({10})^{13 - 12}$

$\sf \longrightarrow \: F = \dfrac{40.2}{40.96} \times10$

$\sf \longrightarrow \: F = \dfrac{4020}{4096} \times10$

$\sf \longrightarrow \: F = \cancel \dfrac{40200}{4096}$

$\sf \longrightarrow \: F = 9.8 \: Newtons$

$\therefore$ The earth exerts a Gravitational force of 9.8 Newtons on a ball of mass 1 kg.

It is due to this large force exerted by the earth that when the 1 kg ball is dropped from a height, it falls to the earth.

Note - Refer to the attachment, if you're facing issues in understanding the calculation part ^^"