Question

How to derive the value of (small) g as :-
$9.8 \: m{s}^{ - 2}$
Step by step explanation. Please.
Don't write if you don't know the correct answer.
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Answer 1

To Derive:-

Value of g on earth's surface. (9.8m/s²)

Derivation:-

•Mass of earth (Mₑ) = 6×10²⁴kg

•Radius of earth (Rₑ) = 6.4×10⁶m

•Value of G = 6.67×10⁻¹¹ Nm²/kg²

Now,we know that:-

=> g = GM/R²

Thus,value of g on earth's surface will be:-

=>gₑ = GMₑ/R²ₑ

=>gₑ = 6.67×10⁻¹¹×6×10²⁴/(6.4×10⁶)²

=>gₑ = 6.67×10⁻¹¹×6×10²⁴/6.4×10⁶×6.4×10⁶

=>gₑ = (6.67×6)/(6.4×6.4)×10⁻¹¹⁺²⁴⁻⁶⁻⁶

=>gₑ = 40.02/40.96×10

=>gₑ = 0.98×10

=>gₑ = 9.8m/s²

Hence,derived !!!

Some Extra Information:-

•The value of g at the surface of earth is 9.8m/s² on an average.

•The value of g decreases with height.

•The value of g decreases with depth.

•The value of g is more at poles and less at equator.

•The value of g is zero at the centre of the earth.