Question

Venus is orbiting around the Sun in 225 days. Calculate the orbital radius and speed of the planet. (Mass of the sun\[ = 2 \times {10^{30}}\,{\text{kg}}\], \[G = 6.67 \times {10^{ - 11}}\,{\text{S}}{\text{.I}}{\text{. unit}}\])

Answer 1

Hint: We can use the expression for Kepler’s third law of planetary motion. Using this formula, we can calculate the radius of the circular orbit of Venus around the Sun. We can use the formula for linear speed of an object in terms of time period. Using this formula, we can calculate the orbital velocity of Venus around the Sun.

Formulae used:
The expression for Kepler’s third law of planetary motion is given by
\[{T^2} = \dfrac{{4{\pi ^2}{R^3}}}{{GM}}\] …… (1)
Here, \[T\] is the time period of revolution, \[R\] is the radius of the circular orbit, \[G\] is universal gravitational constant and \[M\] is the mass of the planet or sun around which the planet or satellite is moving.
The expression for the linear velocity \[v\] of is
\[v = \dfrac{{2\pi R}}{T}\] …… (2)
Here, \[R\] is the radius of the orbit and \[T\] is the time period.

Complete step by step answer:
We have given that the time period of Venus orbiting around the sun is 225 days.
\[T = 225\,{\text{days}}\]
The mass of the Sun is \[2 \times {10^{30}}\,{\text{kg}}\].
\[M = 2 \times {10^{30}}\,{\text{kg}}\]
We have asked to calculate the radius of the orbit of Venus around the Sun and orbital speed of Venus around the Sun. Convert the unit of the time period of Venus around the Sun in the SI system of units.
\[T = \left( {225\,{\text{days}}} \right)\left( {\dfrac{{24\,{\text{h}}}}{{1\,{\text{day}}}}} \right)\left( {\dfrac{{60\,{\text{min}}}}{{1\,{\text{h}}}}} \right)\left( {\dfrac{{60\,{\text{s}}}}{{1\,{\text{min}}}}} \right)\]
\[ \Rightarrow T = 1.944 \times {10^7}{\text{s}}\]
Hence, the time period of Venus is \[1.944 \times {10^7}{\text{s}}\].
We can calculate the radius of orbit of Venus using expression for Kepler’s third law of planetary motion.Rearrange equation (1) for the orbital radius of Venus.
 \[R = {\left( {\dfrac{{GM{T^2}}}{{4{\pi ^2}}}} \right)^{\dfrac{1}{3}}}\]

Substitute \[6.67 \times {10^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2}{\text{/k}}{{\text{g}}^{\text{2}}}\] for \[G\], \[2 \times {10^{30}}\,{\text{kg}}\] for \[M\], \[1.944 \times {10^7}{\text{s}}\] for \[T\] and \[3.14\] for \[\pi \] in the above equation.
\[R = {\left( {\dfrac{{\left( {6.67 \times {{10}^{ - 11}}\,{\text{N}} \cdot {{\text{m}}^2}{\text{/k}}{{\text{g}}^{\text{2}}}} \right)\left( {2 \times {{10}^{30}}\,{\text{kg}}} \right){{\left( {1.944 \times {{10}^7}{\text{s}}} \right)}^2}}}{{4{{\left( {3.14} \right)}^2}}}} \right)^{\dfrac{1}{3}}}\]
\[ \Rightarrow R = 1.08 \times {10^{11}}\,{\text{m}}\]
Hence, the radius of orbit of Venus is \[1.08 \times {10^{11}}\,{\text{m}}\].
Let us now calculate the speed of Venus around the Sun.Substitute \[3.14\] for \[\pi \], \[1.08 \times {10^{11}}\,{\text{m}}\] for \[R\] and \[1.944 \times {10^7}{\text{s}}\] for \[T\] in equation (2).
\[v = \dfrac{{2\left( {3.14} \right)\left( {1.08 \times {{10}^{11}}\,{\text{m}}} \right)}}{{1.944 \times {{10}^7}{\text{s}}}}\]
\[ \therefore v = 34888.9\,{\text{m/s}}\]
Therefore, the orbital speed of Venus around the Sun is \[34888.9\,{\text{m/s}}\].

Hence, the radius and speed of Venus in the orbit around the Sun are \[1.08 \times {10^{11}}\,{\text{m}}\] and \[34888.9\,{\text{m/s}}\] respectively.

Note:We have given the time period of Venus around the Sun with unit days. The students should not forget to convert the unit of time period of Venus around the Sun in the SI system of units because all the physical quantities used in the calculations are with their units in the SI system of units. If one forgets to convert this unit then the final answer will be incorrect.