Gravitational Acceleration Differences in a Pulsar: Calculation and Comparison

  • Context: Graduate 
  • Thread starter Thread starter wcase
  • Start date Start date
  • Tags Tags
    Gravitational
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 6K views
wcase
Messages
9
Reaction score
0
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
 
Astronomy news on Phys.org
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.