Question

Find the orbital period of venus. show work venus: mass=4.82x10^24 orbital radius (m)= 1.08x10^11​

Answer 1

Approximately
`1.94 * 10^(7)\; {\rm s}`.

Step-by-step explanation:

Let
`T` denote the period of this motion.

Assuming that the orbit of the planet venus is circular and centered around the sun, the motion of this planet can be modelled as a centripetal motion.

The net force on the planet would be proportional to the square of its angular velocity
`\omega` where
`\omega = (2\, \pi / T)`. Let
`m` is the mass of the planet and let
`r` denote the radius of the orbit around the sun:

`\displaystyle (\text{net force}) = m\, \omega^(2)\, r = m\, \left((2\,\pi)/(T)\right)^(2)\, r`.

The net force on the planet would also be approximately equal to the gravitational attraction from the sun. Let
`G` denote the gravitational constant, and let
`M` denote the mass of the sun (the solar mass):

`\displaystyle (\text{net force}) = (G\, M\, m)/(r^(2))`.

Equate the two expression for net force to obtain:

`\displaystyle m\, \left((2\,\pi)/(T)\right)^(2)\, r = (\text{net force}) = (G\, M\, m)/(r^(2))`.

Simplify to obtain an expression for
`T`, the period of the orbit:

`\displaystyle T^(2) = ((2\, \pi)^(2)\, r^(3))/(G\, M)`.

`\displaystyle T = (2\, \pi)\, \sqrt{(r^(3))/(G\, M)}`.

Substitute in the following values:

• Gravitational constant
  `G \approx 6.67 * 10^(-11)\; {\rm m^(3)\, kg^(-1)\, s^(-2)}`,
• Solar mass
  `M \approx 1.99 * 10^(30)\; {\rm kg}`, and
• Orbital radius of the planet venus
  `r = 1.08 * 10^(11)\; {\rm m}` (given).

Evaluate to obtain the orbital period of the planet venus under the assumptions:

`\begin{aligned} T &= (2\, \pi)\, \sqrt{(r^(3))/(G\, M)} \\ &\approx (2\, \pi)\, \sqrt{((1.08* 10^(11))^(3))/((6.67 * 10^(-11))\, (1.99 * 10^(30)))}\; {\rm s} \\ &\approx 1.94 * 10^(7)\; {\rm s}\end{aligned}`.