Suppose the orbits of the planets are circles with radii equal to the semimajor

Question

Suppose the orbits of the planets are circles with radii equal to the semimajor axis lengths listed in the Physical Constants sheet on our website, and that the planets are spheres with radii also as given on this page. (a) Make a table of the minimum and maximum distance of each planet from Earth and the corresponding maximum and minimum angular diameter of each planet as seen from Earth, reporting your answer in arcseconds. (Recall: pi radians = 180 degree 60 arcminutes = 1 degree mid 60 arcseconds = 1 arcminute. (b) Atmospheric turbulence usually blurs the images seen through telescopes such that the smallest details that can be discerned are 1-2 arcseconds in diameter. For each of the planets, approximately when in the planet's orbit is it possible to obtain good, detailed pictures of the planet with a ground-based telescope (when is the size of the planet significantly larger than the blurring)?

Explanation / Answer

Hi,

So, In order to get the closest distance between the planets, consider that you have skewed all the planets on a kebab stick. In other words, they are all alligned.

So, now the distance between them is just the difference between their orbital radii.

Example, closest distance between earth and jupiter is abs(14.95-77.78) *10^12 cm = 62.83 * 10^12 cm

Now, to get the maximum distances, let us take the same example of jupiter and earth.

The farthest distance occurs when earth , sun and jupiter all lie on the same line, but the sun is in between earth and jupiter. In other words, we add their radii to get the max distance.

So, the distance here is : (14.95+77.78) *10^12 cm = 92.73 x10^12 cm

Now, about the angular diameters.

If we look at Jupiter, we are seperated by a distance X. Let the planet's radius be R. So, the planet's diameter is 2R.

We can forma a right triangle in which the base is X and the height is 2R. Now, tan(theta) = 2R/X .

If theta is very small, then tan(theta) is approximately equal to theta in radians. To get it in arcseconds, just multiply the answer in radians by 206265 because that is the number of arcseconds in a radian.

To get the max angular diameter, obviously, the planet has to be the closest. So, our X here would be the minimum distance that we calculated.

Similarly, the min angular diameter needs X to be the maximum distance that we calculated.

Plugging in all the values, here are all the answers:

Max distances of the planets from earth, in 10^12 cm : 20.737 ; 25.76 ; 37.729 ; 92.73 ; 157.65 ; 301.87 ; 464.48

Min distances of the planets from earth, in 10^12 cm : 9.163 ; 4.14 ; 7.829 ; 62.83 ; 127.75 ; 271.97 ; 434.58

Max angular diameters, in arcseconds : 109.851 ; 603.03 ; 178.622 ; 469.402 ; 194.616 ; 38.768 ; 23.508

Min angular diameters, in arcseconds : 48.538 ; 96.914 ; 37.064 ; 318.048 ; 157.706 ; 34.928 ; 21.996

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Now, for the second part of the question.

First, we need to consider what can be called as good quality, i.e., since the blur varies from 1-2 arcseconds, the palnet should be atleast 40 arcseconds big for us to see it properly. The factor I have used is 20. Depending on the need, you can change this factor.

Now, the question goes as "....when in the planet's orbit...." .Let us first find the distance from earth to the planets at which the angular diameter is 40 arcseconds.

Mercury, Venus Jupiter and Saturn fulfill this condition always because their minimum angular diameter is always greater than 40 arcseconds.

For the other planets, it is good to observe them when the distance between them and us is atmost :

Mars:1.748*10^13 cm

Uranus: 1.3179*10^14 cm

Neptune: When it is closest (because it barely has its maximum angular diameter above 40 arcseconds)

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