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 percentage error in measurement of acceleration due to gravity on the surface of the earth is n%. The value of n is $$_________________$$

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 percentage error in measurement of acceleration due to gravity on the surface of the earth is n%. The value of n is $$_________________$$

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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$∵ g = \frac{G M}{R^{2}}$
$∴ \frac{Δ g}{g} = \frac{Δ M}{M} + \frac{2 Δ R}{R}$
Percentage error in g is
$\frac{Δ g}{g} \times 100 = \frac{Δ M}{M} \times 100 + \frac{2 Δ R}{R} \times 100 = 1 + 4 = 5 %$

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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

$∵ g = \frac{G M}{R^{2}}$
$∴ \frac{Δ g}{g} = \frac{Δ M}{M} + \frac{2 Δ R}{R}$
Percentage error in g is
$\frac{Δ g}{g} \times 100 = \frac{Δ M}{M} \times 100 + \frac{2 Δ R}{R} \times 100 = 1 + 4 = 5 %$