
Suppose there existed a planet that went around the sun as twice as fast as the earth. What would be its orbital size as compared to that of the earth?
Answer
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Hint: If a planet moves around the sun in an orbit, the gravitational pull of sun to the planet provides the necessary centripetal force.
Complete step by step solution:
Let mass of the planet be ${M_p}$
Distance of the planet from the sun be ${R_p}$
Mass of sun be ${M_s}$
Mass of earth be ${M_e}$
Distance of earth from the sun be ${R_e}$
Gravitational pull between sun and the planet is given by \[\dfrac{{G{M_p}{M_s}}}{{{R_p}^2}}\]
Where G is the universal gravitational constant. The universal gravitational constant is equal to the force of attraction acting between two bodies each of unit mass whose centers are placed unit distance apart.
This gravitational pull provides the necessary centripetal force to the planet to move around the sun.
\[ \Rightarrow \dfrac{{G{M_p}{M_s}}}{{{R_p}^2}} = \dfrac{{{M_p}{v_p}^2}}{{{R_p}}} \Rightarrow \dfrac{{G{M_s}}}{{{R_p}}} = {v_p}^2 \Rightarrow {v_p} = \sqrt {\dfrac{{G{M_s}}}{{{R_p}}}} \]
Gravitational pull between sun and the earth is given by the equation \[\dfrac{{G{M_e}{M_s}}}{{{R_e}^2}}\]
This gravitational pull provides the necessary centripetal force to earth to move around the sun.
\[ \Rightarrow \dfrac{{G{M_e}{M_s}}}{{{R_e}^2}} = \dfrac{{{M_e}{v_e}^2}}{{{R_e}}} \Rightarrow \dfrac{{G{M_s}}}{{{R_e}}} = {v_e}^2 \Rightarrow {v_e} = \sqrt {\dfrac{{G{M_s}}}{{{R_e}}}} \]
It is given that${v_p} = 2{v_e}$
\[\
\Rightarrow \sqrt {\dfrac{{G{M_s}}}{{{R_p}}}} = 2\sqrt {\dfrac{{G{M_s}}}{{{R_e}}}} \Rightarrow \dfrac{1}{{{R_p}}} = \dfrac{2}{{{R_e}}} \Rightarrow {R_p} = \dfrac{{{R_e}}}{2} \\
\\
\]
Hence, the distance of the planet is half that of the distance of the earth from the sun.
Note: A planet revolves around the sun in an elliptical orbit under the influence of gravitational pull of the sun on the planet. This pull or force acts along the line joining the centers of the sun and the planet and is directed towards the sun. The speed by which a planet moves around the sun does not depend on the mass of the planet. It only depends on the distance from the sun.
Complete step by step solution:
Let mass of the planet be ${M_p}$
Distance of the planet from the sun be ${R_p}$
Mass of sun be ${M_s}$
Mass of earth be ${M_e}$
Distance of earth from the sun be ${R_e}$
Gravitational pull between sun and the planet is given by \[\dfrac{{G{M_p}{M_s}}}{{{R_p}^2}}\]
Where G is the universal gravitational constant. The universal gravitational constant is equal to the force of attraction acting between two bodies each of unit mass whose centers are placed unit distance apart.
This gravitational pull provides the necessary centripetal force to the planet to move around the sun.
\[ \Rightarrow \dfrac{{G{M_p}{M_s}}}{{{R_p}^2}} = \dfrac{{{M_p}{v_p}^2}}{{{R_p}}} \Rightarrow \dfrac{{G{M_s}}}{{{R_p}}} = {v_p}^2 \Rightarrow {v_p} = \sqrt {\dfrac{{G{M_s}}}{{{R_p}}}} \]
Gravitational pull between sun and the earth is given by the equation \[\dfrac{{G{M_e}{M_s}}}{{{R_e}^2}}\]
This gravitational pull provides the necessary centripetal force to earth to move around the sun.
\[ \Rightarrow \dfrac{{G{M_e}{M_s}}}{{{R_e}^2}} = \dfrac{{{M_e}{v_e}^2}}{{{R_e}}} \Rightarrow \dfrac{{G{M_s}}}{{{R_e}}} = {v_e}^2 \Rightarrow {v_e} = \sqrt {\dfrac{{G{M_s}}}{{{R_e}}}} \]
It is given that${v_p} = 2{v_e}$
\[\
\Rightarrow \sqrt {\dfrac{{G{M_s}}}{{{R_p}}}} = 2\sqrt {\dfrac{{G{M_s}}}{{{R_e}}}} \Rightarrow \dfrac{1}{{{R_p}}} = \dfrac{2}{{{R_e}}} \Rightarrow {R_p} = \dfrac{{{R_e}}}{2} \\
\\
\]
Hence, the distance of the planet is half that of the distance of the earth from the sun.
Note: A planet revolves around the sun in an elliptical orbit under the influence of gravitational pull of the sun on the planet. This pull or force acts along the line joining the centers of the sun and the planet and is directed towards the sun. The speed by which a planet moves around the sun does not depend on the mass of the planet. It only depends on the distance from the sun.
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