At an average opposition, the Earth and Mars are separated by 0.52 AU. Suppose an astronomer observes Mars at opposition and that seeing blurs the images to a resolution of 1.0 seconds of arc. What is the smallest surface feature the astronomer would be able to resolve on Mars? How does this size compare with the diameter of Mars? (Hint: you will need the small angle equation to solve.)
377km
To determine the smallest surface feature the astronomer would be able to resolve on Mars, we can use the small angle equation:
θ = α / D
Where:
θ is the angular size of the object
α is the linear size of the object
D is the distance to the object
Given:
Distance between Earth and Mars at opposition = 0.52 AU
Resolution of the telescope = 1.0 seconds of arc
First, we need to convert the distance between Earth and Mars into the same units as the resolution of the telescope. 1 AU is approximately equal to 1.496 x 10^8 km.
0.52 AU * 1.496 x 10^8 km/AU = 7.7952 x 10^7 km
Now, we can use the small angle equation to calculate the angular size of the surface feature:
θ = 1.0 seconds of arc = (α / D)
α = θ * D
α = (1.0 seconds of arc) * (7.7952 x 10^7 km)
Note: We assume that the surface feature is relatively small compared to the diameter of Mars, so we can treat the distance between Earth and Mars as the distance to the surface feature.
Using the above calculation, we can determine the linear size (α) of the smallest surface feature the astronomer would be able to resolve.
Now let's compare the size of the surface feature (α) with the diameter of Mars. The diameter of Mars is approximately 6,779 km.
If the calculated linear size (α) is smaller than the diameter of Mars, then the astronomer would be able to resolve the surface feature. If it's larger, the surface feature would be too small to be resolved.
I hope this explanation helps you understand how to approach and solve this problem.