Question 378: Gravitation

Question

The radius of the earth and the radius of orbit around the sun are 6371 km and $149 	imes 10^6 \mathrm{~km}$ respectively. The order of magnitude of the diameter of the orbit is greater than that of earth by

Accepted Answer

Given, radius of earth, $R_e=6371 \mathrm{~km}$ and radius of orbit around the sun, $R_0=149 	imes 10^6 \mathrm{~km}$
Now, the ratio of the diameters of the earth and orbit around the sun can be expressed as
$$\frac{D_0}{D_	heta}=\frac{2 R_0}{2 R_	heta}=\frac{2 	imes 14.9 	imes 10^6}{2 	imes 6371}=2.33 	imes 10^4$$
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 $10^4$.

Answer

$10^3$

Answer

$10^2$

Answer

$10^4$

Answer

$10^5$

Gravitation

The radius of the earth and the radius of orbit around the sun are 6371 km and 149 \times 10^6 \mathrm{~km} respectively. The order of magnitude of the diameter of the orbit is greater than that of earth by

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