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The planet Neptune travels around the Sun with a period of 165 years. Show that the radius of its orbit is approximately thirty times that of Earth's orbit, both being considered as circular.

Note by Rishabh Kumar
2 years, 2 months ago

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Using Kepler third law:

$$\frac{a(neptune)^{3}}{T(neptune)²} = \frac{a(earth)^{3}}{T(earth)²}$$

Where $$a$$ and $$T$$ are respectively radius (generally, half of ellipse greatest axis) and period of trajectory.

Thus:

$$\frac{a(neptune)}{a(earth)} = \frac{T(neptune)}{T(earth)}^\frac{2}{3} = 165^\frac{2}{3}$$

$$\frac{a(neptune)}{a(earth)} = 30.082$$ · 2 years, 2 months ago

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oh in order for the neptune to remain in the orbit its centripetal force should be balanced by the centrifugal force, mv^2 /r = gMm/r^2 v=sq rt gMm/r time period = 2pi R/ velocity from the above time period = 2 pi R/sq rt gmM/r T = 2pi sq rt R^3 /GM R= 165 ^2/3 that of the radius of that earth R= 30 THAT OF EARTH · 2 years, 2 months ago

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