Question

1. A spaceship is moving away from earth with speed ? = 0.6 c. When the ship is at a distance d = 5 ×10^8 km from earth, a radio signal is sent to the ship by the observers on earth. How long does the signal take to reach the ship as measured by the scientist on earth?

2.The kinetic energy and the momentum of a particle deduced from measurements on its track in nuclear photographic emulsions are 250MeV and 368 MeV/c, respectively. Determine the mass of the particle in terms of electron mass? please explain on detail

Answer 1

1] The Lorentz factor for v = 0.6c is:

Let T be the time measured in the Earth's frame when the light reaches the ship.

then, the distance of the ship from the earth at t = T is:

d = [5 x 10¹¹] + (0.6 x 3 x 10⁸)T

where d is in meters

this will be the distance covered by the radio signal in time t.

so, d = cT

=> (3 x 10⁸)T = [5 x 10¹¹] + (0.6 x 3 x 10⁸)T

=> T = 4166.667 s.

this is the time taken in earth's frame.

In the spaceship's frame, the time taken will be:

t = T/1.25 = 3333.33s. [since the time in the earth's frame is larger due to time dilation].