Question

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

Answer 1

Answer:

Option B. 1,100 Earth diameters

Solution:

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

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

$\theta = \frac{arc length(l)}{radius(R)}$                 (1)

where,

From the relation:

l = $\theta \times R$

we get:

distance(d) or R = $\frac{Earth diameter}{\theta}$

distance(d) or R = $\frac{2\times radius of earth}{\theta}$

d = $\frac{2\times 6350000}{8.726\times 10^{-4}}$

distance, d = $1.455\times 10^{10} m$

Comparing it with Earth's diameter:

d = $\frac{1.455\times 10^{10}}{2\times 6350000} = 1,146$

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