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

Option B. 1,100 Earth diameters

Solution:

Angular position of steroid,
`\theta = 0.05^(\circ) = 8.726* 10^(-4) radians` (given)

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

`\theta = (arc length(l))/(radius(R))` (1)

where,

From the relation:

l =
`\theta * R`

we get:

distance(d) or R =
`(Earth diameter)/(\theta)`

distance(d) or R =
`(2* radius of earth)/(\theta)`

d =
`(2* 6350000)/(8.726* 10^(-4))`

distance, d =
`1.455* 10^(10) m`

Comparing it with Earth's diameter:

d =
`(1.455* 10^(10))/(2* 6350000) = 1,146`

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