Calculate the difference in the strength of gravity

In summary, two identical pendulum gravity meters on Mars, one at the equator and one at the North pole, showed a difference in oscillation of 250 times at the equator and 251 times at the pole over a fixed period of time. Using the equation (T(equator)^2)/(T(pole)^2)=(g(equator)^2)/(g(pole)^2), the difference in gravity was calculated to be 0.992. However, this equation is incorrect and should be (g(equator)/g(pole)=sqrt[(T(eq)^2)/(T(pole)^2)], which results in a difference of 0.996. This indicates that the difference in gravity between the equator
  • #1
blueyellow

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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  • #2
I meant g(equator)/g(pole)=sqrt[(250^2)/(251^2)]=0.996
 
  • #3
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?
 
  • #4
I'm in third year, so why isn't that an upper-division course? T is the period
 
  • #5
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.
 
  • #6
but its a course designed for third years. still I have no idea what upper or lower division means
 
  • #7
(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
 
  • #8

1. What is the formula for calculating the difference in the strength of gravity?

The formula for calculating the difference in the strength of gravity is F = G(m1m2)/r², where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between the two objects.

2. How does the distance between two objects affect the strength of gravity?

The strength of gravity is inversely proportional to the square of the distance between two objects. This means that as the distance between two objects increases, the strength of gravity between them decreases.

3. What factors can cause a difference in the strength of gravity?

The strength of gravity between two objects can be affected by their masses and the distance between them. It can also be influenced by other factors such as the gravitational pull of other nearby objects.

4. How can I measure the strength of gravity between two objects?

The strength of gravity can be measured using a device called a gravimeter, which measures the gravitational acceleration between two objects. This is typically done by dropping objects from a certain height and measuring the time it takes for them to fall.

5. How is the strength of gravity different on different planets?

The strength of gravity on different planets varies depending on their mass and size. The larger the planet, the stronger the gravity. For example, the strength of gravity on Earth is about 9.8 m/s² while on Jupiter it is about 24.8 m/s².

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