Calculate the angular velocity of the neptune (orbital period of 165 earth period) in its orbit around the sun.

1 answer:

7 0

The correct answer is: Angular velocity = $1.208 * 10^{-9}$ rad/s

Explanation:
The angular velocity is given as:
ω = $\frac{2\pi}{T}$ --- (1)

Where T = 165 * (365 days) * (24 hours/day) * (60 minutes/hour) * (60 seconds/minute) = 5203440000 s

Plug in the value in (1):
ω = $\frac{2\pi}{5203440000} = 1.208 * 10^{-9}$ rad/s

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A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.959 g, q = 5.84 µC is locat

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Answer:

$8.66\times 10^{-6}\ C$ or $8.66\ \mu C$ .

Explanation:

<u>Given:</u>

Charge on the particle at origin = Q.
Mass of the moving charged particle, $\rm m = 0.959\ g = 0.959\times 10^{-3}\ kg.$
Charge on the moving charged particle, $\rm q = 5.84\ \mu C = 5.84\times 10^{-6}\ C.$
Distance of the moving charged particle from first at t = 0 time, $\rm r=20.7\ cm = 0.207\ m.$
Speed of the moving particle, $\rm v = 47.9\ m/s.$

For the moving particle to circular motion, the electrostatic force between the two must be balanced by the centripetal force on the moving particle.

The electrostatic force on the moving particle due to the charge Q at origin is given by Coulomb's law as:

$\rm F_e = \dfrac{kqQ}{r^2}.$

where, $\rm k$ is the Coulomb's constant having value $\rm 9\times 10^9\ Nm^2/C^2.$

The centripetal force on the moving particle due to particle at origin is given as:

$\rm F_c = \dfrac{mv^2}{r}.$

For the two forces to be balanced,

$\rm F_e = F_c\\\dfrac{kqQ}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow Q = \dfrac{mv^2}{r}\times \dfrac{r^2}{kq}\\=\dfrac{mv^2r}{kq}\\=\dfrac{(0.959\times 10^{-3})\times (47.9)^2\times (0.207)}{(9\times 10^9)\times (5.84\times 10^{-6})}\\=8.66\times 10^{-6}\ C\\=8.66\ \mu C.$