Calculate the difference in the strength of gravity

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Homework Statement



Two spacecraft carrying identical pendulum gravity meters have landed on Mars, one at the equator and one at the North pole. Over a fixed period of time the pendulum at the equator is observed to oscillate 250 times compared with 251 times at the pole. Calculate the difference in the strength of gravity between the equator and the pole. Assuming Mars to be a spherically symmetric, rotating body, derive an expression for the difference in the strength of gravity at the Martian pole and equator due to the rotation of the planet and calculate its value. State what conclusions you draw when comparing the measured difference in gravity between the Martian equator and pole with the difference due to rotation.

The Attempt at a Solution



(T(equator)^2)/(T(pole)^2)=(g(equator)^2)/(g(pole)^2)

(250^2)/(251^2)=g(equator)/g(pole)

0.992=g(equator) /g(pole)
 
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I meant g(equator)/g(pole)=sqrt[(250^2)/(251^2)]=0.996
 
This question doesn't belong in the advanced physics forum, which is for problems from upper-division courses and higher. It should be in the introductory physics forum.

Where did you get that equation from? What does T stand for?
 
I'm in third year, so why isn't that an upper-division course? T is the period
 
You can take lower-division courses during any year. The idea is that the advanced physics forum is for problems of higher difficulty, typical of a junior-level, senior-level, or graduate physics course.

Where did you get the equation from? Did you derive it? It's incorrect.
 
but its a course designed for third years. still I have no idea what upper or lower division means
 
(T(equator)^2)/(T(pole)^2)=(g(equator)^2)/(g(pole)^2)
so

g(equator)/g(pole)=sqrt[(T(eq)^2)/(T(pole)^2)]
g(equator)/g(pole)=sqrt[(250^2)/(251^2)]

how is (T(equator)^2)/(T(pole)^2)=(g(equator)^2)/(g(pole)^2)
wrong? I found it in the notes