ASTR 2020 Space Astronomy
Homework #2
Solutions
1] Orbits: Consider an asteroid orbiting the Sun at a distance of 4 Astronomical Units. The
Earth orbits the Sun with a speed of 30 km/s.
[a] Does this asteroid orbit the Sun faster or slower than the Earth orbits the Sun? [2 points]
Slower
[b] What is the orbital speed of this asteroid around the Sun in km/sec?
[6 points]
Vorbit = (GM / r)0.5 Thus, if Earth move with 30 km/s at 1 AU, at a distance 4 times greater,
any object will orbit with a speed 4 1/2 = 2 times slowe than at 1 AU. Thus, in a circular orbit,
the asteroid will have a speed of 15 km s-1.
2] Escape speed from an asteroid: Suppose that you landed on a spherical asteroid with a radius
of 1.0 km. It has a mass, M = 1016 grams.
[a] What is the escape speed from the surface of the asteroid in meters per second?
[10 points]
Vesc = (2GM / r)0.5 = (2 [6.67x10-8]x[1016] / [105])0.5 = 115 cm s-1 = 1.15 m s-1
[b] Would you be able to hit a baseball or golf ball with sufficient speed to escape the
gravitational field of this asteroid?
[2 points]
Yes
[c] Would you have to be careful not to jump too hard or run too fast on the surface? Why or
why not?
[4 points]
Yes. You could jump or run off the asteroid because your speed could easily exceed the
escape velocity from the asteroid !
[d] What is the density of this asteroid in grams cm-3?
Density = ρ = M / (4 π / 3) r3 = 2.387 = 2.4 g cm-3.
[10 points]
3] Doppler Effect 1: The Earth orbits the Sun with a velocity of 30 km/s. Suppose you were
living on a spaceship far-far away that was located in the orbital plane of the Earth. You and the
spaceship are at rest with respect to the Sun. Suppose someone on earth sent you a radio
message with a transmitter frequency f = 100 MHz ( f = 108 Hz).
[a] When the Earth is moving directly toward you, would you receive a lower or higher
frequency? Why?
[4 points]
Higher frequency. Because the source and you are approaching each other so the
frequency increases.
[b] What would be the frequency that you received?
[10 points]
8
6
10
Δf / f = V/c Δf = f V/c = [10 Hz][3x10 cm/s] / [2.998x10 ] = 10,006.67 Hz ~ 10 kHz
freceived = f + Δf ~ 1.0001 x 108 Hz
[c] Suppose you had a friend in another spaceship located far-far way, and directly above the
orbital plane of the Solar system. He is at rest with respect to the Sun. What would be the
frequency of the radio transmission when he receives it from Earth?
[2 points]
No change in frequency => freceived = 108 Hz because the source is moving at right angles.
4] Light travel time: Suppose in the above example, you are about 1018 cm from the Earth.
How much time would elapse between the transmission of the radio signal, and its reception?
Δt = x / c = [1018 cm] / [3 x 1010 cm s-1] = 3 x 107 seconds ~ 1 year.
[10 points]
5] Doppler Effect 2: Your spaceship now moves directly away from the Solar system at 10% of
the speed of light (speed of light is approximately c = 3 x 1010 cm s-1). Someone on Earth shines
a very bright green laser having a wavelength of 0.5 micro-meters (0.5 µm = 5.0 x 10-5 cm).
What is the wavelength of the laser light when you see it?
[10 points]
Δλ / λ = V/c = 0.1 c /c = 0.1 => Δλ= 0.1 λ = 0.05 µm = 5 x 10-6 cm
λ = λ + Δλ = 5.0 x 10-5 + 5.0 x 10-6 = 0.000055 cm = 5.5 x 10-5 cm = 0.55 µm
6] Aberration of starlight:
a) What is meant by the “aberration of starlight”?
[10 points]
The apparent deflection in the position of stars by a small angle in the direction of motion.
For the Earth orbiting the Sun, the angle is
b) Imagine observing a star at right angles to the orbit-plane of the Earth. By how much does the
apparent position of the star change (in arc-seconds) during the course of the year? [10 points]
θ = Vorbit / c (radians) = [Vorbit / c] x 206,265 (arcseconds)
= ([3 x 106 cm/sec] / [ 2.998 x 1010 cm/sec]) x 206,265 = 20.64 ~ 21 arcseconds.