Question

Calculate the value of ‘g’ if G = 6.67 × 10-11 Nm²/kg², mass of the Earth = 6×10²⁴ kg and radius of the

Earth = 6400 km.​

Answer 1

Given Question :

Calculate the value of ‘g’ if G = 6.67 × 10^-11 Nm²/kg², mass of the Earth = 6×10²⁴ kg and radius of the Earth = 6400 km.

Required Solution :

» Given-

• G ( gravitational constant ) = $\sf { 6.67 \times {10}^{-11} N{m}^{2}}$

• m ( mass of the earth ) = $\sf {6 \times {10}^{24}kg}$

• r (Radius of the earth) = 6400 km

Converting to its SI unit-

$\rm { \leadsto 1 km = 1000m}$

$\rm { \leadsto 6400 km = 64,00,000m}$

$\rm { \leadsto 6400 km = 64 \times {10}^{5}m}$

$\rm\red { \leadsto 6400 km = 6.4 \times {10}^{6}m}$

» To find-

• g (acceleration due to gravity)

» Calculation-

Using the formula-

• $\boxed {\large \tt { g = \dfrac{Gm}{ {r} ^{2} } }}$

Substitute the value to find the value of “g”

$\tt { : \implies g = \dfrac{ 6.67 \times {10}^{-11} \times 6 \times {10}^{24} }{ {{6.4} \times ({10}^{6}) }^{2}} }$

$\qquad$

$\tt { : \implies g = \dfrac{ 6.67 \times {10}^{-11+24} \times 6 }{ {6.4}^{2} \times {(10) }^{6 \times 2}} }$

$\qquad$

$\tt { : \implies g = \dfrac{ 6.67 \times {10}^{13} \times 6 }{ 6.4 \times 6.4 \times {(10) }^{12}} }$

$\qquad$

$\tt { : \implies g = \dfrac{ 6.67 \times {10}^{13-12} \times 6 }{ 6.4 \times 6.4 } }$

$\qquad$

$\tt { : \implies g = \dfrac{ 667 \times 1 \cancel{0} \times 6 \times 1 \cancel {0} }{ 64 \times 64 \times 1 \cancel {00} } }$

$\qquad$

$\tt { : \implies g = \dfrac{ 4002 \times 10}{4096}}$

$\qquad$

$\tt { : \implies g = \cancel { \dfrac{ 40020}{4096}} }$

$\qquad$

$\tt \purple { : \implies g = 9.7m/{s}^{2} \: or \: 9.8m/{s}^{2} }$

Therefore, value of “g” is $\tt{ 9.7m/{s}^{2} \: or \: 9.8m/{s}^{2} }$

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Note : Refer to the attachment if you get any confusion in the calculation terms.

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