Question

Calculate the density of earth and its density error

d=m/v
where m = 6.00*10^24kg
and v = (4/3)(pi)(6359000)^3
= 1.07709...*10^21 m^3

therefore d=6.00*10^24/1.07709...*10^21 m^3
d = 5570.5254...kg/m^3
using the sig fig rule, in division and multiplication take the least decimal value and round the answer to that...
therefore d=5.57*10^3kg/m^3

then for error analysis, i did the below..
since i need the error analysis for the volume i used the formula
deltaV = n(deltaR / R), where r is the radius and n is the coefficient of the power
therefore deltaV = 3(20000/6359000)
delta V = 3.145148...*10^-3
then to calculate the delta D i used the formula:
deltaD = D(sqrt[(deltaM/M)^2 + (deltaV/V)^2])
therefore delta D = 5.57*10^3(sqrt[(5.00*10^22/6.00*10^24)^2/(3.145148*10^-3/1.07709*10^21)^2])
delta D = 46.42104...

I re did this calculations and i am getting a different value, please help me out and check whether it is correct or not

Answer 1

The density calculation looks fine and I would agree with the 3 sig figs in the answer as the mass is to 3 sig figs.


I am not entirely sure what you are doing with the error estimation, however

the delta V seems small

As a crude estimate the error in m is say +/- 0.01x10^24 or 1.7%, which swamps the error in V which is about 0.05%

Thus the error in the density will be dominated by error in the mass, so the error in the density will be +/-

5.57x10^3 x 1.7/100 = 95 kg m^-3

which will be an over estimate.

so your value of 46 is of the correct magnitude