A) 4 days
B) 16 days
C) \[2\sqrt{2}\]days
D) 64 days
Correct Answer: C
Solution :
Key Idea: Parking orbit is a geostationary satellites orbit. From Keplers third law of planetary motion, the square of period of revolution (T) is directly proportional to cube of semi-major axis of its elliptical orbit , i.e., \[{{T}^{2}}\propto {{a}^{3}}\] Given, \[{{T}_{1}}=1\] day (geostationary) \[{{a}_{1}}=a,\,\,{{a}_{2}}=2\,a\] \[\therefore \] \[\frac{{{T}_{1}}^{2}}{{{T}_{2}}^{2}}=\frac{{{a}_{1}}^{3}}{{{a}_{2}}^{3}}\] \[\Rightarrow \] \[{{T}_{2}}^{2}=\frac{{{a}_{2}}^{3}}{{{a}_{1}}^{3}}{{T}_{1}}^{2}=\frac{{{(2\,a)}^{3}}}{{{a}^{3}}}\times 1=8\] \[\Rightarrow \] \[{{T}_{2}}=2\sqrt{2}\] days
You need to login to perform this action.
You will be redirected in
3 sec
