Circular orbits and satellites

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**Gravity** — the inward pull of the central body. There is no separate 'orbit force'.

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Pointing **toward the centre** of the circular path. The centripetal force is whatever points inward and curves the motion into a circle.

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$\dfrac{GMm}{r^{2}} = \dfrac{mv^{2}}{r}$ — the orbiting mass m cancels from both sides.

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$v = \sqrt{\dfrac{GM}{r}}$ — derived from gravity equalling the centripetal force. A bigger radius gives a smaller speed.

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**No** — the mass cancels, so v = √(GM/r) depends only on G, the central mass M and the radius r.

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**Toward the central body** (centripetal) — the same direction as gravity.

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$T^{2} = \dfrac{4\pi^{2}}{GM}\,r^{3}$ — period squared is proportional to radius cubed; the constant 4π²/GM depends only on the central mass M.

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Rearrange Kepler's third law: $M = \dfrac{4\pi^{2} r^{3}}{G\,T^{2}}$ — measure an orbit's period T and radius r.

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A circular orbit with a period of exactly **24 h**, in Earth's spin direction, above the **equator** — so the satellite stays above one fixed point.

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**height = r − (radius of the planet)**, because r is measured from the planet's centre.

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v = √(GM/r) falls as r rises (slower), while T² = (4π²/GM)r³ rises steeply with r (much longer period).

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v = √(GM/r) = √[(6.67×10⁻¹¹ × 6.0×10²⁴) / 7.0×10⁶] ≈ 7.6 × 10³ m s⁻¹.