On the grid of the Hertzsprung–Russell (HR) diagram shown, draw a line to represent the approximate position of the main sequence.

Barnard’s star is a main sequence star that is 1.8 pc from Earth.

(i) Define the parsec.

(ii) Calculate the parallax angle of Barnard’s star as measured from Earth.

Outline, using your answer to (b)(ii) and a labelled diagram, how the distance of Barnard’s star from Earth is measured.

The apparent brightness of Barnard’s star is 3.6×10^–12Wm^–2 and its surface temperature is 3800 K.

Given that 1 pc=3.1×10¹⁶m, show for Barnard’s star

(i) that its luminosity is of the order of 10²³W.

(ii) that its surface area is of the order of 10¹⁶m².

any suitable line from anywhere in top left-hand quadrant; (accept a straight line)
to bottom right-hand quadrant;
The shaded areas are the limits within which the line must be drawn.

(i) distance at which 1 AU subtends an angle of 1 arcsec / distance at which the angle subtended by the radius of Earth’s orbit is 1 arcsec;

(ii) $p = \left( {\frac{1}{d} = } \right)0.56{\mathop{\rm arcsec}\nolimits}$;

Labelled diagram should relate to the following points:
measure against the fixed stars the angle Barnard’s star subtends at Earth in June and again in December;
difference between the two angles is twice the parallax angle;
orbital radius of Earth about Sun is 1 AU so distance to star is computed from $d = \frac{1}{p}$;

(i) L=4πbd²;
=4×3.14×3.6×10^-12×[1.8×3.1]²×10³²;
=1.4×10²³W;
≈10²³W

(ii) $A = \frac{L}{{\sigma {T^4}}}$;
$= \frac{{1.4 \times {{10}^{23}}}}{{5.67 \times {{10}^{ - 8}} \times {{3.8}^4} \times {{10}^{12}}}}$; (allow ECF from (d)(i))
=1.184×10¹⁶m$^2$;
≈10¹⁶m²