Mars is approximately 1.5 A.U. from the Sun. Now imagine yourself living on Mars. How often do you see the Earth in opposition? A) Once every Martian yearB) Once every Earth yearC) Once every...

Mars is approximately 1.5 A.U. from the Sun. Now imagine yourself living on Mars. How often do you see the Earth in opposition?

A) Once every Martian year
B) Once every Earth year
C) Once every 2.19 Earth years
D) Once every 8 months
E) Never

Asked on by sruland

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mathsworkmusic's profile pic

mathsworkmusic | (Level 2) Educator

Posted on

The answer is

C) Once every 2.19 Earth years

The time between events such as two planets in opposition (eg when the Sun is diametrically opposite to Mars viewed from Earth) is called the 'synodic period'.

[Wikipedia:

The synodic period is the temporal interval that it takes for an object to reappear at the same point in relation to two or more other objects, e.g. when the Earth is directly between Mars and the Sun, ie Earth is directly in the way of the view from Mars to the Sun ]

Knowing that Mars is 1.5 AU from the Sun (with the reference distance being Earth at 1 AU from the Sun), we know that Mars has a circumference 1.5 times that of the Earth. However, it doesn't take 1.5 times to go round (1.5 Earth years) since the gravitational pull from the Sun is smaller than for the Earth and thus it is orbiting the Sun at a slower speed than the Earth is.

To work out the orbital velocity relative to Earth's, we use Kepler's Law

V = 1/R^0.5

where V is the relative orbital velocity and R is the relative distance from the Sun.

Given that R=1.5 (approx), this gives V =  0.816. (In actuality R is more like 1.524 and V is more like 0.802, as this is a simplified model).

The length of the Mars year is then R/V = 1.5/0.816 = 1.84 times the Earth year (actually closer to 1.9) since time = distance/velocity.

The synodic period S concerning two celestial bodies satisfies the relation

1/S = 1/T1 - 1/T2

where T1 is the orbit of body 1 and T2 of body 2. Mars and Earth in opposition then occurs every S = 1/(1/T1 - 1/T2) Earth years if T1,T2 are measured in Earth years. Thus S = 1/(1/1 - 1/1.84) = 2.19 Earth years.

A quicker equation to use is S = 1/(1-1.5^(-3/2)) = 2.19 (or S = 2.13 if using R = 1.524 AU, which is a more accurate value)

Answer C) Once every 2.19 Earth years

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caledon's profile pic

caledon | High School Teacher | (Level 3) Senior Educator

Posted on

This is true, but note that the question specifies that you're living on Mars, and it is the Earth that is in opposition, not Mars itself.

By definition, opposition means that two objects (usually a planet and the sun) appear in opposite parts of the sky. Also by this definition, planets inferior to the observer can never be in opposition.

Thus, if we're on Mars, the Earth can E) Never be in opposition because it cannot be on the opposite side of the sky from the sun, in the same way that Mercury and Venus, as viewed from Earth, are always located close to the Sun. From Mars, we would say that the Earth and the Sun are in conjunction every 2.19 years.

mathsworkmusic's profile pic

mathsworkmusic | (Level 2) Educator

Posted on

Yes, it would be 'the transit of Earth' I suppose, but as it is all measured in Earth-relative units it is rather difficult to move to Mars-centric thinking.

Mars in opposition to the Sun from Earth = Earth traversing the Mars skies. When I get to live on Mars, I'll look forward to the transit of the Earth every 2.13 Earth years, if I remember how long an Earth year is by then. More conveniently, I might remember that a Martian year is 1.9 times as long as an Earth year, so I could look forward to a transit of Earth every 2.13/1.9= 1.12 Martian years which might seem like less time if my long-term body clock had adjusted.

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