Question

Explain that the gravitational acceleration value is 9.8.​

Answer 1

Explanation :

A free fall body of mass (m) from a distance (R) From the centre of the earth of mass (M), then the gravitational force between the earth and the body Is :-

$\sf \:g = \frac{GM} {{R}^{2} }$

• Value of G = 6.67 × 10-11Nm²/kg²
• m = 6 × 16²⁴kg
• R = 6.4 × 10⁶m

$\sf \: g = \frac{6 .67 \times {10}^{ - 11} \times 6 \times 10 {}^{24} }{(6.4 \times 10 {}^{6} ) {}^{2} } \\ \\ \sf \: = \frac{6.67 \times 6 \times {10}^{13} }{6.4 \times 6.4 \times {10}^{12} } \\ \\ \sf \: = \frac{6.67 \times 6 \times {10}^{13 - 12} }{6.4 \times 6.4} \\ \\ \sf \: = \frac{6.67 \times 6 \times 10}{6.4 \times 6.4} \\ \\ \: = \red{\boxed{ \sf \: 9.8m / {sec}^{2} }}$

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Answer 2

Explanation :

A free fall body of mass (m) from a distance (R) From the centre of the earth of mass (M), then the gravitational force between the earth and the body Is :-

\begin{gathered}\sf \:g = \frac{GM} {{R}^{2} } \\\end{gathered}

g=

R

2

GM

Value of G = 6.67 × 10-11Nm²/kg²

m = 6 × 16²⁴kg

R = 6.4 × 10⁶m

\begin{gathered}\sf \: g = \frac{6 .67 \times {10}^{ - 11} \times 6 \times 10 {}^{24} }{(6.4 \times 10 {}^{6} ) {}^{2} } \\ \\ \sf \: = \frac{6.67 \times 6 \times {10}^{13} }{6.4 \times 6.4 \times {10}^{12} } \\ \\ \sf \: = \frac{6.67 \times 6 \times {10}^{13 - 12} }{6.4 \times 6.4} \\ \\ \sf \: = \frac{6.67 \times 6 \times 10}{6.4 \times 6.4} \\ \\ \: = \red{\boxed{ \sf \: 9.8m / {sec}^{2} }}\end{gathered}

g=

(6.4×10

6

)

2

6.67×10

−11

×6×10

24

=

6.4×6.4×10

12

6.67×6×10

13

=

6.4×6.4

6.67×6×10

13−12

=

6.4×6.4

6.67×6×10

=

9.8m/sec

2

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