Mean solar day is the time interval between two successive noons when the sun passes through the zenith point (meridian). $A$ sidereal day is the time interval between two successive transits of a distant star through the zenith point (meridian). By drawing an appropriate diagram showing the earth's spin and orbital motion,show that the mean solar day is $4\,\text{min}$ longer than the sidereal day. In other words,distant stars would rise $4\,\text{min}$ early every successive day.

(N/A) Every day,the Earth advances in its orbit by approximately $1^{\circ}$.
To have the sun at the zenith point again,the Earth must rotate by $361^{\circ}$ (which we define as $1$ mean solar day).
Since $361^{\circ}$ corresponds to $24\,\text{h}$,we can calculate the time taken for $1^{\circ}$ as follows:
$\because \text{To cover } 361^{\circ}, \text{time taken} = 24\,\text{h}$
$\because \text{To cover } 1^{\circ}, \text{time taken} = t$
$t = \frac{24}{361} \times 1 \approx 0.0664\,\text{h}$
$t = 0.0664 \times 60\,\text{min} \approx 3.99\,\text{min} \approx 4\,\text{min}$.
Hence,the mean solar day is $4\,\text{min}$ longer than the sidereal day,and distant stars rise $4\,\text{min}$ early every successive day.