Questions on the apparent magnitude scale

Question 1

State which way the apparent magnitude scale runs, and the apparent magnitude of the dimmest stars visible to the naked eye.

Accepted Answer

Brighter objects have smaller, and then negative, magnitudes; the scale runs backwards (1). The dimmest naked-eye stars are about magnitude +6 (1).

Question 2

Define the parsec, and state its relationship to the light year.

Accepted Answer

One parsec is the distance at which one astronomical unit, the mean Earth-Sun distance, subtends an angle of one arcsecond (1). 1 pc = 3.26 light years (1).

Question 3

State the brightness ratio corresponding to one step of apparent magnitude, and to five steps, and explain why the scale is built from ratios rather than equal differences.

Accepted Answer

One magnitude step is a received-intensity ratio of 2.51 (1); five steps are exactly 100 (1). The eye judges brightness roughly logarithmically, so the equal-feeling steps of the ancient scale correspond to equal ratios of intensity (1).

Question 4

Define the astronomical unit and the light year.

Accepted Answer

The astronomical unit is the mean distance from the Earth to the Sun (1). The light year is the distance light travels through a vacuum in one year (1).

Question 5

A star lies exactly 10 pc from Earth. State the relationship between its apparent and absolute magnitudes.

Accepted Answer

They are equal: at d = 10 pc, m − M = 5 log(10/10) = 0 (1).

Question 6

State what is meant by the distance modulus of a star, and what it is used to find.

Accepted Answer

The difference m − M between the apparent and absolute magnitudes (1). Since m − M = 5 log(d/10), it gives the star's distance in parsecs (1).

Question 7

Star P has apparent magnitude 2.0 and star Q has apparent magnitude 7.0. Calculate how many times brighter P appears than Q.

Accepted Answer

P is the brighter, having the smaller number, by 2.51^(7.0 − 2.0) = 2.51⁵ (1) = 100 times (1)

Question 8

Sirius has apparent magnitude −1.5 and lies 2.64 pc from Earth. Calculate its absolute magnitude.

Accepted Answer

M = m − 5 log(d/10) (1) = −1.5 − 5 log(0.264) (1) M = +1.4 (1)

Question 9

A star has absolute magnitude +0.6 and lies 25 pc away. Calculate its apparent magnitude.

Accepted Answer

m = M + 5 log(d/10) (1) = 0.6 + 5 log(2.5) (1) m = +2.6 (1)

Question 10

A star's apparent magnitude is 8.5 and its absolute magnitude is 1.0. Calculate its distance in parsecs.

Accepted Answer

m − M = 7.5 = 5 log(d/10) (1) log(d/10) = 1.5 (1) d = 10 × 10^1.5 (1) d = 316 pc, about 3.2 × 10² pc (1)

Question 11

Two stars have the same absolute magnitude. One lies 5.0 pc away, the other 50 pc. Calculate the difference in their apparent magnitudes and the ratio of their apparent brightnesses.

Accepted Answer

m differs by 5 log(50/5.0) (1) = 5.0 magnitudes (1) The nearer star appears 2.51⁵ = 100 times brighter (1)

Question 12

A star radiates a total power (luminosity) of 3.2 × 10²⁸ W. Calculate the radiant flux received by a detector 4.1 × 10¹⁷ m away. Assume no light is absorbed on the way.

Accepted Answer

F = L/4πd² (1) = 3.2 × 10²⁸/(4π × (4.1 × 10¹⁷)²) (1) = 1.5 × 10⁻⁸ W m⁻² (1).

Question 13

A pulsating star used as a standard candle has a known luminosity of 7.8 × 10²⁹ W. The radiant flux received from it is 6.2 × 10⁻¹³ W m⁻². Determine its distance in metres, and in parsecs (1 pc = 3.08 × 10¹⁶ m).

Accepted Answer

F = L/4πd², so d = √(L/4πF) (1) = √(7.8 × 10²⁹/(4π × 6.2 × 10⁻¹³)) (1) = 3.2 × 10²⁰ m (1) = 3.2 × 10²⁰/(3.08 × 10¹⁶) ≈ 1.0 × 10⁴ pc (1).

Question 14

Star V appears 16 times brighter than star W. Calculate the difference between their apparent magnitudes.

Accepted Answer

2.51^Δm = 16, so Δm = log 16/log 2.51 (1) = 3.0 magnitudes, V carrying the smaller number (1).

Question 15

A star lies 42 light years away and has apparent magnitude 5.0. Calculate its distance in parsecs and its absolute magnitude.

Accepted Answer

d = 42/3.26 (1) d = 12.9 pc (1) M = m − 5 log(d/10) = 5.0 − 5 log(12.9/10) (1) M = +4.4 (1)

Question 16

The Sun's apparent magnitude is −26.7 and Sirius's is −1.5. Estimate how many times brighter the Sun appears than Sirius.

Accepted Answer

The magnitude difference is 25.2 (1) Ratio = 2.51^25.2 (1) ≈ 1.2 × 10¹⁰ (1)

Question 17

Explain why apparent magnitude alone cannot tell you how luminous a star really is, and what absolute magnitude does about the problem.

Accepted Answer

Apparent magnitude mixes the star's true output with its distance: a feeble nearby star can look brighter than a distant giant (1). Absolute magnitude removes the distance by quoting the apparent magnitude every star would have at the same standard 10 pc (1).

Question 18

A red dwarf has absolute magnitude +7.5 and lies 3.5 pc from Earth. The naked-eye limit is apparent magnitude +6. Deduce whether the star is visible to the naked eye.

Accepted Answer

m = M + 5 log(d/10) (1) = 7.5 + 5 log(0.35) = 7.5 − 2.3 (1) = +5.2 (1). This is a smaller (brighter) magnitude than the +6 limit, so the star is just visible to the naked eye (1).

Question 19

Star T has absolute magnitude −2.0 and lies 250 pc away. Star U has absolute magnitude +1.0 and lies 15 pc away. Deduce which star appears brighter in the night sky.

Accepted Answer

T: m = −2.0 + 5 log(25) = +5.0 (1). U: m = +1.0 + 5 log(1.5) = +1.9 (1). U has the smaller apparent magnitude (1), so U appears the brighter, even though T is by far the more luminous star (1).

Question 20

The mean Earth-Sun distance is 1.50 × 10¹¹ m, and one arcsecond is 1/3600 of a degree. Show that one parsec is about 3.1 × 10¹⁶ m.

Accepted Answer

1 arcsecond = (1/3600) × (π/180) = 4.85 × 10⁻⁶ rad (1). One parsec is the distance at which 1 AU subtends this angle, so d = 1.50 × 10¹¹/(4.85 × 10⁻⁶) (1) = 3.09 × 10¹⁶ m ≈ 3.1 × 10¹⁶ m (1).