Question

The time period of Jupiter is \[11.6\text{ years}\] . How far is Jupiter from the Sun? Distance of the earth from the Sun is \[\text{1}\text{.5}\times \text{1}{{\text{0}}^{\text{11}}}m\] .
(A) \[{{r}_{J}}=7.68\times \text{1}{{\text{0}}^{\text{11}}}m\]
(B) \[{{r}_{J}}=9\times \text{1}{{\text{0}}^{\text{11}}}m\]
(C) \[{{r}_{J}}=68\times \text{1}{{\text{0}}^{\text{11}}}m\]
(D) \[{{r}_{J}}=8\times \text{1}{{\text{0}}^{\text{11}}}m\]

Answer 1

Hint:Fifth in line from the Sun, Jupiter is the largest planet in the solar system: more than twice as massive as all the other planets combined. Jupiter's has characteristic stripes and swirls, which are cold, windy clouds of ammonia and water, floating in an atmosphere of hydrogen and helium. Jupiter’s Great Red Spot is a giant storm bigger than the size of Earth that has raged for centuries. To find the relation between the time period and the distance of a planet from the sun, we can make use of Kepler’s Laws.

Formula Used:\[{{T}^{2}}\propto {{a}^{3}}\]

Complete step by step solution: Kepler’s third law of planetary motion is known as the law of Periods.  It states that the square of the time period of the planet is directly proportional to the cube of the semi-major axis of its orbit, that is \[{{T}^{2}}\propto {{a}^{3}}\] .
Since for practical applications to celestial bodies, we consider the orbits to be roughly circular, we can replace the value of the semi-major axis with the radius of the orbit, which is the distance of the planet from the Sun.
The modified Kepler’s Law can be given as \[{{T}^{2}}=k\times {{r}^{3}}\] where \[k\] is the constant of proportionality and \[r\] is the distance of the planet from the sun.
Firstly, we’ll consider the equation for Earth. Substituting the value of the time period and the distance of the earth from the sun, we can say that
\[{{(1\text{ year)}}^{\text{2}}}\text{=k}\times {{(\text{1}\text{.5}\times \text{1}{{\text{0}}^{\text{11}}})}^{3}}\,-------(1)\]
Now, we can substitute the values for Jupiter and find another equation as follows
\[{{(11.6\text{ year)}}^{\text{2}}}\text{=k}\times {{({{r}_{J}})}^{3}}\,--------(2)\]
In the above equation, \[{{r}_{J}}\] denotes the distance of Jupiter from the sun.
Dividing the two equations obtained above, we get
\[{{\left( \dfrac{1\text{ year}}{11.6\text{ year}} \right)}^{2}}={{\left( \dfrac{\text{1}\text{.5}\times \text{1}{{\text{0}}^{\text{11}}}}{{{r}_{J}}} \right)}^{3}}\]
Rearranging the terms and adjusting the indices of the terms, we get
\[\begin{align}
  & {{r}_{J}}=(\text{1}\text{.5}\times \text{1}{{\text{0}}^{\text{11}}}m)\times {{\left( \dfrac{11.6\text{ year}}{1\text{ year}} \right)}^{\dfrac{2}{3}}} \\
 & \Rightarrow {{r}_{J}}=(\text{1}\text{.5}\times \text{1}{{\text{0}}^{\text{11}}}m)\times {{\left( 134.56 \right)}^{\dfrac{1}{3}}} \\
\end{align}\]
Further solving the above equation, we can say that
\[\begin{align}
  & {{r}_{J}}=(\text{1}\text{.5}\times \text{1}{{\text{0}}^{\text{11}}}m)\times (5.16) \\
 & \Rightarrow {{r}_{J}}=7.74\times \text{1}{{\text{0}}^{\text{11}}}m \\
 & \Rightarrow {{r}_{J}}\approx 7.68\times \text{1}{{\text{0}}^{\text{11}}}m \\
\end{align}\]

Hence, we can say that option (A) is the correct answer.

Note: Students often make mistakes in calculating the values of terms raised to fractional indices. The safe way to do such calculations is to first raise the term to the power of the numerator and then consider the denominator, as shown in the solution. This method reduces the chances of errors. Also note that we do not always obtain an accurate answer and we might have to round off or approximate to the closest answer, as we did with the final value.