Orbital radius of mercury around the sun is 0.38 A.U. Find maximum elo

1.	Orbital radius of mercury around the sun is 0.38 A.U. Find maximum elongation for mercury and its distance from earth at this elongation.
Answer» We have r_PS = 0.38 AU Using r_PS = AU sin θ, or sin θ = \(\frac{r_{PS}}{AU}\) = sin^-1 0.38 = 22^o3’ Now r_PE = AU cos θ ∴ r_PE = (1.496 × 10¹¹ m) × cos 22^o3’ = 1.384 × 10¹¹ m = 1.384 × 10⁸ km.

Orbital radius of mercury around the sun is 0.38 A.U. Find maximum elongation for mercury and its distance from earth at this elongation.

Answer»

We have

r_PS = 0.38 AU

Using r_PS = AU sin θ,

or sin θ = \(\frac{r_{PS}}{AU}\)

= sin^-1 0.38 = 22^o3’

Now r_PE = AU cos θ

∴ r_PE = (1.496 × 10¹¹ m) × cos 22^o3’

= 1.384 × 10¹¹ m

= 1.384 × 10⁸ km.