Acceleration due to gravity on the surface of the earth is $$g=GMR2$$. The gravitational constant G is exactly known. But percentage error in measurement of the mass of earth M and radius of the earth R are $$1$$% and $$2$$%. Respectively. The maximum percent age error is measurement of acceleration due to gravity on the surface of the earth in n%. Find the value of n.

Question

Acceleration due to gravity on the surface of the earth is $$g=GMR2$$. The gravitational constant G is exactly known. But percentage error in measurement of the mass of earth M and radius of the earth R are $$1$$% and $$2$$%. Respectively. The maximum percent age error is measurement of acceleration due to gravity on the surface of the earth in n%. Find the value of n.

Infinity Learn · Accepted Answer

∵    $$g=GMR2$$∴     Δ$$gg=$$Δ$$MM+2$$ΔRRPercentage error in g isΔgg×$$100=$$ΔMM×$$100+2$$ΔRR×$$100$$             $$=1+4=5$$%

Accuracy; precision of instruments and errors in measurements

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$\begin{array}{l} ∵ g = \frac{GM}{R^{2}} \\ ∴ \frac{Δg}{g} = \frac{ΔM}{M} + \frac{2 ΔR}{R} \end{array}$
Percentage error in g is
$\begin{matrix} \frac{Δg}{g} \times 100 = \frac{ΔM}{M} \times 100 + \frac{2 ΔR}{R} \times 100 \\ = 1 + 4 = 5 % \end{matrix}$

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Accuracy; precision of instruments and errors in measurements

Remember concepts with our Masterclasses.

∵ g=GMR2∴ Δgg=ΔMM+2ΔRRPercentage error in g isΔgg×100=ΔMM×100+2ΔRR×100 =1+4=5%

Ready to Test Your Skills?

free study material

$\begin{array}{l} ∵ g = \frac{GM}{R^{2}} \\ ∴ \frac{Δg}{g} = \frac{ΔM}{M} + \frac{2 ΔR}{R} \end{array}$
Percentage error in g is
$\begin{matrix} \frac{Δg}{g} \times 100 = \frac{ΔM}{M} \times 100 + \frac{2 ΔR}{R} \times 100 \\ = 1 + 4 = 5 % \end{matrix}$