# Video: Variation of Objects Weights with Different Gravitational Accelerations

Ed Burdette

A body of mass 1.00 × 10² kg was weighed, using a spring scale, at the north pole and at the equator. Assuming 𝑔 = 9.83 m/s² at the north pole and 𝑔 = 9.78 m/s² at the equator, what was the difference in the two readings?

02:21

### Video Transcript

A body of mass 1.00 times ten to the two kilograms was weighed using a spring scale, at the North Pole and at the equator. Assuming π equals 9.83 metres per second squared at the North Pole and π equals 9.78 metres per second squared at the equator, what was the difference in the two readings?

Weβll call this difference π₯π€. Also weβll call the mass of the body 1.00 times ten to the two kilograms m. Weβll call the acceleration due to gravity at the North Pole 9.83 metres per second squared π sub π. And weβll call the acceleration due to gravity at the equator 9.78 metres per second squared π sub π.

If we draw a sketch of the two places where the body was weighed at the North Pole, we know the acceleration due to gravity was π sub π and at the equator the acceleration was π sub π. Recall that the weight of an object is equal to the mass of that object times the acceleration due to gravity.

In our case, we want to solve not for weight, but for the change in weight or difference in weight. π₯π€, that difference, is equal to the mass of the object which does not change times the change in the acceleration due to gravity, π₯π. We can rewrite π₯π as π sub π minus π sub π.

Since the problem statement tells us m, π sub π, and π sub π, we can now insert these values into this equation. The difference in the objects weight in the two readings is 1.00 times ten to the two kilograms times 9.83 minus 9.78 metres per second squared. Entering these values on our calculator, we find a number of 5.00 newtons. To three significant figures, thatβs the difference in the weight of the two readings.