UCCS PES 1060: Astronomy II
PROBLEMS CH 17 Spring 2015
1. The distance to a star in parsecs is:
1
.
parallax
name: __________________________
Spica’s parallax is 0.0120
For Spica, its distance is:
1
 83
0.012
pc.
2. Neptune orbits the Sun at 30.1 AU, about 30 times farther from the Sun than Earth.
Thus, the baseline for Neptune would be 30.1 times greater, and the parallax angle would be that many times larger.
Find the parallax angle:
(parallax of Spica) * (Distance from Sun to Neptune) =
0.36 ".
3. A star lying 20 pc from the Sun has proper motion of 0.50/yr.
Find the actual transverse distance the star travels in one year, using the angular distance formula:
1 pc
1 ly
1 AU
=
=
=
3.2616 ly
9.4905 x 1015 m
149,597,900,000 m
Convert 1 pc into A.U.’s
(1 pc)*( 3.2616 ly/pc )*( 9.4905 x 1015 m/ly )*( 1 AU / 149,597,900,000 m ) =
actual distance traveled
A.U.
= (star distance from the sun) [( proper motion per year /
)]
 0.5 
9
pc
20 
  0.000048 pc  1.5 10
206265


actual distance traveled =
x (3.0856 x 1013 km/pc) =
4. transverse velocity
=
seconds in one year
=
1.5 109
 47.5
3.16 107
=
sec.
km/sec
5. Get the line-of-sight component of the velocity using the Doppler Effect.
If the spectral lines are shifted by:
0.01 %, then the line-of-sight velocity is: 0.01 % of the speed of light,
(0.01) / (100) =
0.0001
*
(speed of light in km) = 30
6. Use the Pythagorean Theorem to find the combination of these two velocities:
v 2  (47.5)2   30   3156.25 ,
2
v2 = ( answer from # 4
v2 =
v=
v=
)
2
which gives: v = 56 for the space velocity.
+ (
+
)
2
=
=
√
km/sec
for the space velocity.
km/sec.