A) \[{{r}_{1}}^{2/5}\]
B) \[{{({{r}_{1}}+{{r}_{2}})}^{3/2}}\]
C) \[{{({{r}_{1}}-{{r}_{2}})}^{3/2}}\]
D) \[{{r}_{1}}^{3/2}\]
Correct Answer: B
Solution :
In astronomy the point of closest approach is called the pericentre and that of farthest excursion is called apocentre: When a is semi major axis we have Periapsis \[=(1-e)a\] Apoapsis = (1 + e) a From condition given in question, \[{{r}_{1}}=a(1+e)\] \[{{r}_{2}}=a(1-e)\] \[\Rightarrow \] \[{{r}_{1}}+{{r}_{2}}=2a\] \[\Rightarrow \] \[a=\frac{{{r}_{1}}+{{r}_{2}}}{2}\] Also from Kepler's third law of planetary motion we have, the square of the period of revolution (T) of any planet around the sun is directly proportional to the cube of the semi major axis of its elliptic orbit . i.e., \[{{T}^{2}}\propto {{a}^{3}}\] \[\Rightarrow \] \[T={{a}^{3/2}}\] \[\propto {{\left( \frac{{{r}_{1}}+{{r}_{2}}}{2} \right)}^{3/2}}\] \[\Rightarrow \] \[T\propto {{({{r}_{1}}+{{r}_{2}})}^{3/2}}\]
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