Using just the facts that the period of earth around the sun is 1 year. that the distance of 1 degree of latitude on earth is 100km . that the acceleration due to gravity on earth is 10m/s^2 and that the angular diameter of the sun from earth is 0.5 degrees, is there a way to calculate the earth-sun distance?

This question is prompted by a question I was given in class. the actual question ask you to find the density ratio between the sun and the earth and i've managed to calculate all the info need for the earth.

Now if I can just figure out how to calculate the earth-sun distance I can use that to calculate the mass by rearranging T and the volume by using trigonometry to find the radius of the sun and then plugging it in.

The thing is I have looked all over for some hints and It seems that all the methods ( the ones i could find anyway) use some third body like the moon in their calculations. I want to do with no additional info to what I mentioned.

Also do you think it's possible that I could just write that the distance is 1au which has accepted value ..... do you think that still counts as only using the accepted data( as in is it a basic enough value that i can just plug it in) or is there a better way to find it mathematically ?

  • $\begingroup$ Since 1 AU is defined as the distance of the Sun to the Earth, that answer would be circular. Can you use stellar parallax? $\endgroup$
    – Schwern
    Commented Oct 1, 2018 at 23:20
  • $\begingroup$ @Schwern how could I go about using that to find the distance ? $\endgroup$
    – bhapi
    Commented Oct 1, 2018 at 23:43
  • $\begingroup$ @Schwern I thought that paralax relied on the astronomical units .... wouldn't arguing with it become circular aswell ? $\endgroup$
    – bhapi
    Commented Oct 1, 2018 at 23:48
  • $\begingroup$ @Schwern I'm probably wrong, that's just the reason I hadn't used it because I don't quite understand how to go about it... $\endgroup$
    – bhapi
    Commented Oct 1, 2018 at 23:58
  • $\begingroup$ You can combine the period of Earth's revolution around the sun with the mass of the sun to get an absolute distance between the Earth and the Sun (assuming circular orbit). Or you can calculate the distance using the angular size of the Sun and the physical size of the sun. I don't believe you can do it with the information provided only, but I may be missing some clever way. $\endgroup$
    – enumaris
    Commented Oct 2, 2018 at 0:01

1 Answer 1


You've got...

  • Orbital period of 1 year
  • 1° = 100km
  • g = 10m/s^2
  • angular diameter of the Sun = 0.5°

You have the Sun's angular diameter. Angular diameter is the ratio of the distance to its diameter: $2 arctan(\frac{diameter_{sun}}{2 distance_{sun}})$ If we also knew the Sun's diameter we could calculate its distance. But we don't.

You have the orbital period of the Earth. That depends on the mass of the Sun, the radius of the Earth's orbit, and the Gravitational Constant. $t = \frac{2 \pi r^3}{GM_{sun}}$. If we have two of those we can solve for the third. The Gravitational Constant can be determined by direct measurement, so that just leaves the mass of the Sun. Historically the mass of the Sun was worked out from the radius of Earth's orbit, not vice-versa.

1° = 100km means assuming the Earth is a perfect sphere the circumference of the Earth is 36,000 km (it's actually 1° = 111km and 40,000 km). That means its radius is $\frac{C_e}{2\pi}$ or 5732 km (6371 km in reality).

And here we're stuck. We'd need additional observations to work it out.

It turns out we can use simple trigonometry to determine the distance from the Earth to the Sun.

Place two observers on opposite sides of the Earth. Have them both measure the angle to the Sun at exactly the same time when the Sun is perfectly perpendicular to the surface of the Earth, 90°. After compensating for axial tilt, the other will get a slightly different angle.

These two observers, plus the Sun, form a right triangle. The distance between the observers is the base. The angle is the observed angle. The hypotenuse is the distance to the Sun. $cos(angle) = \frac{adjacent}{hypotenuse}$. Solving for the hypotenuse: $hypotenuse = \frac{adjacent}{cos(angle)}$. Adjacent is the diameter of the Earth, we have that. And we have an angle. So we can work out the hypotenuse, or distance to the Sun.

This difference in angle is very, very, very small. It's about 0.0046° or about 16 arcseconds.

In modern day with a universal time keeping system it's relatively easy to coordinate such observations. Back in the 18th century such accurate time keeping was impossible. This is why they instead measured the transit of Venus from many points on the globe. Again using these observations and simple geometry they could determine the distance to the Sun.

  • 2
    $\begingroup$ Maybe it will be useful to make clear, as a conclusion, that given only the information in the OP you cannot determine the Sun-Earth distance. The degree of latitude and value of g are irrelevant anyway. The Earth could have different size and different surface gravity and still be at the same distance from the Sun. $\endgroup$
    – nasu
    Commented Oct 2, 2018 at 1:45
  • $\begingroup$ @schwern when you say the observed angle do you mean the angular diameter ? $\endgroup$
    – bhapi
    Commented Oct 2, 2018 at 12:36
  • $\begingroup$ I tried to work out the earth sun distance from what you said but This is what I got. b=diameter of earth=36000km. observed angle =0.5degrees which gives the hypotenuse as 36001370.82 km but this is far from the known value of 149.6miliion km. Am i misunderstanding the method you suggest ? $\endgroup$
    – bhapi
    Commented Oct 2, 2018 at 12:47
  • $\begingroup$ @nasu is there some way to express the distance in terms of the earths mass or some other trick. all i actually needis to show the ratio of density between the sun and earth given this data $\endgroup$
    – bhapi
    Commented Oct 2, 2018 at 12:49
  • $\begingroup$ It may be useful if you post the actual problem you are trying to solve. As it was given to you. Maybe you started on a path which takes you astray. $\endgroup$
    – nasu
    Commented Oct 2, 2018 at 14:51

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