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.

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

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TopNewestUsing 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:

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

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

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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

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