A teacher uses a buzzer attached to a string to demonstrate the Doppler effect to a group of students - Scottish Highers Physics - Question 5 - 2022

To find the redshift (z) of star B, we use the formula:

$z = \frac{\lambda_{observed} - \lambda_{rest}}{\lambda_{rest}}$

Given:

• Wavelength from the sun, (\lambda_{rest} = 580 ; nm)
• Wavelength from star B, (\lambda_{observed} = 610 ; nm)

Substituting the values:

$z = \frac{610 - 580}{580} = \frac{30}{580} \approx 0.05172$

Thus, the redshift of star B is approximately 0.052.

Using the redshift (z) calculated previously:

The relationship between redshift and distance can be approximated using:

$D = \frac{c}{H_0} z$

where:

• (c = 3.00 \times 10^8 ; m/s)
• Assuming (H_0 \approx 70 ; km/s/Mpc \approx 2.3 \times 10^{-18} ; s^{-1})

Substituting the values:

$D = \frac{3.00 \times 10^8}{2.3 \times 10^{-18} \times 0.052} \approx 6.8 \times 10^6 \; m$

This provides an approximate distance of 6.8 million meters from Earth to the binary star system.

The gravitational force (F) between two masses is given by Newton's law of gravitation:

$F = G \frac{m_1 m_2}{r^2}$

Where:

• (G = 6.67 \times 10^{-11} ; N m^2/kg^2)
• (m_1 = 2.19 \times 10^{30} ; kg)
• (m_2 = 1.80 \times 10^{30} ; kg)
• (r = 3.44 \times 10^{12} ; m)

Substituting:

$F = 6.67 \times 10^{-11} \frac{(2.19 \times 10^{30})(1.80 \times 10^{30})}{(3.44 \times 10^{12})^2}$

Calculating:

$F \approx 2.22 \times 10^{13} \; N$

At this point, the distance between the stars is halved. The gravitational force is inversely proportional to the square of the distance:

$F \propto \frac{1}{r^2}$

Thus, if the distance is halved, the force increases by a factor of:

$F_{new} = 4 \times F_{old}$

The gravitational force between star A and star B is four times greater at this point compared to that in (i).