110H AST - Practical Astronomy University of Toronto Mississauga Assignment 1 Solution
1. Orion, shown on the map below, is a prominent constellation in the winter sky. Make a cross sta and measure the angular separation of stars and in Orions belt.
If the cross piece of the cross sta is 1 cm, then the angular separation between and would be matched by positioning the cross piece down the cross sta about 41.5 cm from your eye. The angle is then 1 cm 2.4096 102 radians. 41.5 cm To make this angle more understandable, it should be converted to degrees, using the fact that 360 = 2 radians = 6.2832 radians. Therefore, 2.4096 102 radians 360 = 1.3806 . 6.2832 radians
Note, the uncertainty in reading the x values is probably 0.1 cm. We can try computing the angle with x = 41.4 cm and x = 41.6 cm. This shows that we are not justied carrying more signicant gures than we have in our angle. The angle can then be converted into minutes of arc and to seconds of arc to get 1.3806 = 1 22 50 . For comparison, the actual angular separation is 1.3879 . 2. For your answer in question 1, what error is made by using the usual small angle approximation instead of the exact trigonometric equation? To use an exact trigonometric equation, we divide the triangle of the cross sta down the middle to create a right triangle. The length down the right triangle is still 41.5 cm, but we use half the size of the cross piece, 0.5 cm. Then the equation is tan 2 = 0.5 cm . 41.5 cm
Solving for the angular separation, , gives = 2 arctan 0.5 cm = 1.3806 . 41.5 cm
We nd that, to the number of signicant gures used here, there is no error produced by using the approximation instead of the exact equation. Stating this another way, using the small angle approximation produces an error < 1 for angles of a degree or less. 3. The Moons angular diameter is 0 31 36 at its average separation from Earth (384,404 km). Because of the Moons elliptical orbit, this separation varies by (1 moon ) the average separation. Here, moon is the eccentricity of the Moons orbit, and its value is moon = 0.055. Could the variation in the Moons separation be measured with a cross sta? Justify your answer. Using the Moons average separation from Earth and the Moons angular diameter at that time enables us to calculate the Moons actual diameter. First convert the Moons average angular diameter from degrees to radian, 0 31 36 = 0 31.6 = = 0.526666 9.192 103 radians. Then, the Moons actual diameter is DMoon = 9.192 103 radians 384, 404 km = 3533.4 km. This values is also given on page 21 of the course note. Next, use the Moons average separation from Earth and the eccentricity of the Moons orbit, to nd the separation between the Earth and Moon at the closest and farthest separations. farthest separation = (1.0 + 0.055) 384, 404 km = 405, 546 km,
and closest separation = (1.0 0.055) 384, 404 km = 363, 262 km. Then, combine the Moons actual diameter and the two separations to nd the angular sizes of the Moon at the two separations. 3533.4 km = 9.72698 103 radians = 0 33 26 . at closest separation = 363, 262 km at farthest separation = 3533.4 km = 8.71280 103 radians = 0 29 57 . 405, 546 km
The dierence between the two angular sizes is 1.01418 103 radians = 3 29 . Because our eye can distinguish angles down to about 1 , we might just be able to see this angular dierence. 4. The city of Toronto has a diameter of about 50 km. If Earths atmosphere blurs angles smaller than 1 , is it possible to see a feature the size of Toronto on the surface of Mars tonight, when the distance between Mars and Earth is 0.7 AU? First, convert the distance to Mars from AU to km, d = 0.7 AU (1.496 108 km) = 1.047 108 km. Then calculate the angular size of an area the diameter of Toronto, 50 km = = 4.7746 107 radians 0.1 . 1.047 108 km Because the blurring of Earths atmosph...