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

The redshift ( ext{z}) is calculated using the formula:

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

Given: ( ext{\lambda_{observed}} = 610, nm) and ( ext{\lambda_{rest}} = 580, nm):

$z = \frac{610 - 580}{580} = \frac{30}{580} \approx 0.0517$

Using the redshift formula along with the relationship of distance and redshift, the distance (D) can be calculated from the redshift:

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

Where (c = 3.00 \times 10^8 m/s) and assuming a Hubble constant (H_0) for calculation, we find:

If (z \approx 0.0517):

D \approx 0.0517 \cdot \frac{3.00\times10^8}{67.8 \times 10^3} \approx 2.3 \times 10^8 m\

The gravitational force (F) is calculated using Newton's law of gravitation:

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

Where:

• (G = 6.67 \times 10^{-11} \frac{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 these values yields:

$F = 6.67 \times 10^{-11} \frac{(2.19 \times 10^{30})(1.80 \times 10^{30})}{(3.44 \times 10^{12})^2} \approx 2.22 \times 10^{20} N$

If the distance between the stars is halved, the gravitational force increases by a factor of (4). This is because gravitational force is inversely proportional to the square of the distance: (F \propto \frac{1}{r^2}), so halving the distance results in: (4 \times F_{original}).