The radius of the earth and the radius of orbit around the sun are 6371 km and `149 xx 10^(6)` km respectively. The order of magnitude of the diameter of the orbit is greater than that of earth by

Options

A. `10^(3)`
B. `10^(2)`
C. `10^(4)`
D. `10^(5)`

Answer

C

Explanation

Given, radius of earth, `R_(e)` = 6371 km and radius of orbit around the sun, `R_(0) =149 xx 10^(6)` km
Now, the ratio of the diameters of the earth and orbit around the sun can be expressed as
`(D_(0))/(D_(e))=(2R_(0))/(2R_(e))=(2xx14.9xx10^(6))/(2xx6371)=2.33xx10^(4)`
From above ratio, it can be concluded that the order of magnitude of the diameter of the orbit is greater than that of earth by multiple of `10^(4)` .