Gravitational Acceleration Differences in a Pulsar: Calculation and Comparison

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The discussion focuses on calculating the difference between gravitational acceleration (ag) and free-fall acceleration (g) for a pulsar with specific mass and radius. The equations used are ag = GM/R^2 and g = ag - w^2R, leading to values of ag and g that are significantly different. The user seeks clarification on how to express the percentage difference, considering the correct formula for percentage difference as ((g - ag) / g) * 100. Ultimately, the conclusion is that the correct calculation for the percentage difference should be based on the relationship between g and ag, confirming that the approach of using ((ag - g) / ag) * 100 is valid. The thread emphasizes the importance of precise calculations in physics problems.
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Here is the question:

Consider a pulsar, a collapsed star of extremely high density, with a mass M equal to that of the Sun (1.98 × 1030 kg), a radius R of only 12.7 km, and a rotational period T of 0.0545 s. By what percentage does the free-fall acceleration g differ from the gravitational acceleration ag at the equator of this spherical star?

Just to make sure I am doing it right, I have the equations
ag=GM/R^2
g=ag-w2R

and when I plug in the values I get ag=8.1881*1017
g=8.18642*1011

From what I gather from my textbook, ag is the gravitational acceleration and g is the free-fall.

Now my predicament is that this is homework submitted online, I have a certain amount of submissions, and I only have one remaining.

So when I compare the percentages, should I do ag/g*100=1.0002*108%
or g/ag*100=.0001%

Thanks
 
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your problem mentions differ so I think you need to calculate a difference ( (g - ag) / g) *100 but that's just my guess.
 
that makes sense, but now I have to choose from ((g-ag)/g)*100=-1.0002e8 and
((ag-g)/ag)*100=99.9999

but I guess since subtraction isn't reversible, it is most likely ((g-ag)/g)
 
In the above problem accelaration increases so (ag-g)/g*100 is the correct answer
 
"(ag-g)/g*100 is the correct answer"

Indeed, thanks.
 
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