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Planet X has 20 times the mass of the earth and four times the earth's radius. It orbits star Y...

Question:

Planet {eq}X {/eq} has {eq}20 {/eq} times the mass of the earth and four times the earth's radius. It orbits star {eq}Y {/eq} at a distance of {eq}18 \ AU {/eq}, where {eq}1 \ AU {/eq} is the earth-sun distance. Star {eq}Y {/eq} has twice the mass of our sun.

a) What is the orbital period of Planet {eq}X {/eq} in years?

b) An astronaut has an earth weight of {eq}140 \ lbs {/eq}. What is her weight on the surface of planet {eq}X {/eq}?

Kepler's Law:

This law explains the movement of planets in space. Three fundamentals laws of the Kepler applied for the planets. It is used for measuring the time period corresponding to the radius and mass of the planet.

Answer and Explanation:

Given data

  • The mass of planet X is {eq}{M_X} = 20{M_e} = 20 \times 5.972 \times {10^{24}}\;{\rm{kg}} {/eq}.
  • The radius of planet X is {eq}{R_X} = 4{R_e} {/eq}.
  • The orbital radius of planet X around star Y is {eq}{R_X} = 18\;{\rm{AU}} = 18 \times 1.496 \times {10^{11}}\;{\rm{m}} {/eq}.
  • The mass of star Y is {eq}{M_Y} = 2{M_{{\rm{sun}}}} = 2 \times 1.989 \times {10^{30}}\;{\rm{kg}} {/eq}.


(a)

The expression for the orbital period using Kepler's law is,

{eq}{T_X} = \sqrt {\dfrac{{4{\pi ^2}{R_X}}}{{G{M_Y}}}} {/eq}

Here, the universal gravitational constant is {eq}G = 6.67 \times {10^{ - 11}}\;{{\rm{m}}^3}{\rm{k}}{{\rm{g}}^{ - 1}}{{\rm{s}}^{ - 2}} {/eq}.


Substitute values.

{eq}\begin{align*} {T_X} &= \sqrt {\dfrac{{4{\pi ^2}{{\left( {18 \times 1.496 \times {{10}^{11}}\;{\rm{m}}} \right)}^3}}}{{\left( {6.67 \times {{10}^{ - 11}}\;{{\rm{m}}^3}{\rm{k}}{{\rm{g}}^{ - 1}}{{\rm{s}}^{ - 2}}} \right)\left( {2 \times 1.989 \times {{10}^{30}}\;{\rm{kg}}} \right)}}} \\ &= \sqrt {2.905 \times {{10}^{18}}\;{{\rm{s}}^{\rm{2}}}} \\ &= 1.704 \times {10^9}\;{\rm{s}} \times \dfrac{{3.17098 \times {{10}^{ - 8}}\;{\rm{yr}}}}{{1\;{\rm{s}}}}\\ &= 54.046\;{\rm{yr}} \end{align*} {/eq}


Thus, the orbital period of X is 54.046 years.


(b)

The expression for the gravity of earth is,

{eq}{g_e} = \dfrac{{G{M_e}}}{{R_e^2}} {/eq}


The gravity of planet X is,

{eq}\begin{align*} {g_X} &= \dfrac{{G{M_X}}}{{R_X^2}}\\ &= \dfrac{{G\left( {20{M_e}} \right)}}{{{{\left( {4{R_e}} \right)}^2}}}\\ &= \dfrac{{20}}{{16}}\dfrac{{G{M_e}}}{{R_e^2}}\\ &= 1.25{g_e} \end{align*} {/eq}


  • The weight of astronaut on earth is {eq}{W_e} = 140\;{\rm{lbs}} {/eq}.


The weight of astronaut on planet X will be,

{eq}\begin{align*} {W_X} &= m{g_X}\\ &= 1.25m{g_e}\\ &= 1.25{W_e}\\ &= 1.25\left( {140\;{\rm{lbs}}} \right)\\ &= 175\;{\rm{lbs}} \end{align*} {/eq}


Thus, the weight of astronaut on planet X is 175 lbs.


Learn more about this topic:

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Kepler's Three Laws of Planetary Motion

from Basics of Astronomy

Chapter 22 / Lesson 12
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