A comet passes Earth over its equator. Pictures are taken of it from Earth's north and south poles simultaneously. In these pictures, the position of the asteroid differs by 0.05°. How far away was the asteroid from Earth when the pictures were taken?

A. 0.00087 Earth diameters

B. 1,100 Earth diameters

C. 45,000 Earth diameters

D. 4,100,000 Earth diameters

Respuesta :

Answer:

Option B. 1,100 Earth diameters

Solution:

Angular position of steroid, [tex]\theta = 0.05^{\circ} = 8.726\times 10^{-4} radians[/tex]                           (given)

To calculate the distance of asteroid, we use parallax method given as:

[tex]\theta = \frac{arc length(l)}{radius(R)}[/tex]                 (1)

where,

From the relation:

l = [tex]\theta \times R[/tex]

we get:

distance(d) or R = [tex]\frac{Earth diameter}{\theta}[/tex]

distance(d) or R = [tex]\frac{2\times radius of earth}{\theta}[/tex]

d = [tex]\frac{2\times 6350000}{8.726\times 10^{-4}}[/tex]

distance, d = [tex]1.455\times 10^{10} m[/tex]

Comparing it with Earth's diameter:

d = [tex]\frac{1.455\times 10^{10}}{2\times 6350000} = 1,146[/tex]

Since, the value is close to 1,100 Earth diameters, therefore, option B is the right answer.

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