Why is the weight of an object greater on than on the ?
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Gravitational Forces in 2D
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Recall that an object's weight is the gravitational force exerted on it by Earth, which depends on the gravitational acceleration \(g\) at that location. The weight \(W\) is given by \(W = mg\), where \(m\) is the object's mass and \(g\) is the local acceleration due to gravity.
Understand that the acceleration due to gravity \(g\) varies with altitude and latitude. It decreases with increasing altitude because the object is farther from Earth's center, and it varies with latitude due to Earth's rotation and its oblate shape (flattened at the poles and bulging at the equator).
Compare the given locations:
- At 10 km above the Equator (in an airplane), the altitude is high, so \(g\) is less than at sea level.
- At the top of Mount Everest, the altitude is about 8.8 km, so \(g\) is also less than at sea level.
- At sea level at the Equator, the altitude is lowest, but due to Earth's equatorial bulge and rotation, \(g\) is slightly less than at the poles.
- At sea level at the North Pole, the altitude is low, and because the Earth is flattened at the poles and rotation effects are minimal, \(g\) is greatest here.
Use the formula for gravitational acceleration considering Earth's radius \(R\) and altitude \(h\):
\(g = G \frac{M}{(R + h)^2}\),
where \(G\) is the gravitational constant and \(M\) is Earth's mass. Since \(h\) is smallest at sea level, \(g\) is larger there compared to higher altitudes.
Also consider the effect of Earth's rotation, which reduces the effective gravity more at the Equator than at the poles. This means the effective weight is greatest at the poles, especially at sea level, making the object's weight greatest at sea level at the North Pole.
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