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Nata [24]
3 years ago
12

A comet passes Earth over its equator. Pictures are taken of it from Earth's north and south poles simultaneously. In these pict

ures, 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
Physics
1 answer:
jok3333 [9.3K]3 years ago
4 0

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.

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Explanation:

Let's suppose the material in question is a spring with spring constant k, mass m and position k, the kinetic energy possessed by the string will be;

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