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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.

#Physics #Mechanics

Easy Math Editor

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Comments

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

yaswanth thod - 6 years, 5 months ago

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:

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

$\frac{a(neptune)}{a(earth)} = 30.082$

mat baluch - 6 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$