Why is the gravitational field strength halfway up Mount Everest the same as at sea level at the equator?I'm trying to understand Newton's Laws and gravitational field strength.
The key to answering this question is to understand that the earth is not a perfect sphere but rather an oblate spheroid. That means that its diameter is relatively larger around the equator than the diameter measured from pole to pole. The difference in diameters is about 25 miles, whereas Mt. Everest is about 6 miles high.
Now the universal law of gravitation does state that the force is proportional to the mass and inversely proportional to the square of the distance between two objects.
However, because the earth is deformed, the distance from the center of the earth to Mt. Everest is about the same as the distance from the center of the earth to the surface of the sea at the equator. Therefore, the gravitational field strength is the same at both points.
It is because the earth is not a perfect sphere. You already know, if you have been studying this topic, that gravity varies depending on distance from the center of the gravitational pull (1/R^2). If the earth were a perfect sphere, the gravitational pull would be less if you were halfway up a high mountain like Everest than at sea level; not much, but measurable. But the earth is an oblate spheroid; it is "flattened" a bit, as if two hands had pressed at the north and south poles. So the slightly lesser gravitational pull at the equator caused by the earth being slightly wider there (and therefore, that the surface is farther from the earth's core) is balanced out by the distance halfway up Everest.