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Why is it inaccurate to use mgy to calculate the potential energy of a satellite orbiting earth at a height one earth radius above the earth's surface

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Final answer:

The formula U = mgy is not accurate for calculating the gravitational potential energy of a satellite orbiting at a significant height because g is not constant at such distances. Instead, the formula U = -GMm/r should be used. This reflects the gravitational potential becoming more negative at larger distances, which means orbits at those distances are bound due to the total energy remaining negative.

Step-by-step explanation:

It's inaccurate to use the simple formula U = mgy to calculate the gravitational potential energy of a satellite orbiting Earth at a significant height because this formula assumes that the gravitational acceleration (g) is constant. However, g varies with altitude and is less at higher altitudes. For satellites in orbit, especially at a height of one Earth radius above the surface, we need to use a more general expression that accounts for the change in g with distance from the center of Earth.

The correct formula for calculating the gravitational potential energy (GPE) of a satellite at a distance r from the center of Earth (where r is significantly greater than the Earth's radius) is derived from the law of universal gravitation and is U = -GMm/r, where G is the gravitational constant, M is the mass of Earth, and m is the mass of the satellite.

This more general expression shows that at large distances, GPE becomes increasingly negative, approaching zero as r becomes infinitely large. A satellite with negative total energy is in a bound orbit because the kinetic energy (which is always positive) is not enough to overcome the magnitude of the negative GPE, hence the satellite remains gravitationally bound to Earth. As for the trend in the ratio of kinetic energy to the change in potential energy as the size of the orbit increases, it generally decreases, indicating that for larger orbits, the kinetic energy constitutes a larger fraction of the total energy needed to maintain the orbit.

The assumption made in this derivation is that the satellite orbits around the center of the astronomical body (such as Earth), and not around a common center of mass, which is not exact but works well since Earth is much more massive compared to the satellite, as in the case of the Moon orbiting the Earth.

User Soycharliente
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The equation for the force of gravity is F=Gm(planet)m(satellite)/radius(of orbit)^2


Therefore, the square of the distance between the objects (the earth and the satellite) inversely affects the force of gravity on the satellite. This means that the farther apart the objects are, the smaller the force of gravity is (not the constant acceleration of 9.8m/s^2).
User Sasha Tsukanov
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