Planet Distance

Question 1

A classic radiotelecope detects a planet’s hydrogen emitted waves of $\lambda=21 cm$. If the signal emitted by a planet has a wavelength that is 0.28 cm longer than the standard 21 cm wavelength what is the distance to this planet? (Instead of planet there could be a galaxy for example)

The redshift is given by the equation

$z +1= \lambda(obs)/\lambda_0 =21.28/21 =1.01333$

$z =0.01333$

The speed is thus

$1+z = \sqrt{(1+v/c)/(1-v/c)}$

$1.02684 =(1+x)/)(1-x)$

$1.02684 -1.02684*x =1+x$

$0.02684 =2.02684*x$

$x =0.01324$

$v =0.01324*c =3972.7 km/s$

The relation between speed and distance is

$V = H*d$

where H is the Hubble constant $H =67.80 km/(s*Mpc)$ Mpc is MegaParsec

Therefore the distance d to the planet is

$d = V/H = 3972.7/67.80 =58.594 Mpc =58.594*3.27 MLy =191.11 MLy$ (Mega Light years)

Question 2

Two cars move from oposite directions one towards the other.  The figure gives their speed dependence on time.  The vertical is vs=42.0 m/s. If the drivers start to decrease their speed  when the cars are at a distance of 201 from each other, what is the minimum distance between cars? The slope of the first velocity graphic (solid line) is $a1=-42/5 m/s^2$ . It means the motion equation of first car is (space as function of time):

$x1(t) = 0+Vs*t + a1*t^2/2$

The slope of the second velocity graphic is (dashed line) (in module) $a2= 31.5/4 m/s^2$. This shows that the equation of motion of the second car is

$x2(t) = 201 -31.5*t + a2*t^2/2$

When both cars are stopped thei speed is zero: $V1(t1)=V2(t2)=0$.

First car is stopped after 5 seconds (we deduce this from the graphic)

$x1(5) = 42*5 -42/5 *5^2/2 =105 m$

The second train stops after 4 seconds. This is the x intersection point of the graphic

$x2(4) = 201-31.5*4 +(31.5/4)*(4^2/2) =138m$

Thus the trains separation is $\Delta(x) = x2-x1 =138-105 =33 m$