Question

Problem 6.37

In recent years, astronomers have found planets orbiting nearby stars that are quite different from planets in our solar system. Kepler-12b, has a diameter that is 1.7 times that of Jupiter (R_Jupiter = 6.99

Answer 1

Concepts and reason

The concepts used to solve this problem are acceleration due to gravity and solar system.

Calculate the acceleration due to gravity on the size of Kepler-12b planet.

Fundamentals

The expression for the acceleration due to gravity of the planet is,

$g = \frac{{Gm}}{{{r^2}}}$

Here, $g$ is the acceleration due to gravity of the planet, G is the gravitational constant, m is the mass of the planet, and r is the radius of the planet.

The acceleration due to gravity of the planet is,

$g = \frac{{Gm}}{{{r^2}}}$

Rewrite the expression in terms of acceleration due to gravity of Kepler-12b planet.

$g' = \frac{{G\left( {0.43{m_J}} \right)}}{{{{\left( {1.7{r_J}} \right)}^2}}}$

Here, $g'$ is the acceleration due to gravity of Kepler-12b planet, G is the gravitational constant, ${m_J}$ is the mass of the Jupiter planet, and ${r_J}$ is the means radius of the Jupiter planet.

The expression for the acceleration due to gravity of Kepler-12b planet is,

$g' = \frac{{G\left( {0.43{m_J}} \right)}}{{{{\left( {1.7{r_J}} \right)}^2}}}$

Substitute $6.67 \times {10^{ - 11}}\,{{\rm{m}}^3} \cdot {\rm{k}}{{\rm{g}}^{ - 1}} \cdot {{\rm{s}}^{ - 2}}$ for $G$ , $1.90 \times {10^{27}}\,{\rm{kg}}$ for ${m_J}$ , and $6.99 \times {10^7}\,{\rm{m}}$ for ${r_J}$ to find $g'$ .

$\begin{array}{c}\\g' = \frac{{\left( {6.67 \times {{10}^{ - 11}}\,{{\rm{m}}^3} \cdot {\rm{k}}{{\rm{g}}^{ - 1}} \cdot {{\rm{s}}^{ - 2}}} \right)\left( {0.43} \right)\left( {1.90 \times {{10}^{27}}\,{\rm{kg}}} \right)}}{{{{\left[ {\left( {1.7} \right)\left( {6.99 \times {{10}^7}\,{\rm{m}}} \right)} \right]}^2}}}\\\\ = 3.86\,{\rm{m/}}{{\rm{s}}^2}\\\end{array}$

Therefore, the acceleration due to gravity of Kepler-12b planet is $3.86\,{\rm{m/}}{{\rm{s}}^2}$ .

Ans:

The acceleration due to gravity of Kepler-12b planet is $3.86\,{\rm{m/}}{{\rm{s}}^2}$ .