Past exam of the physics 1 paper of the University of Oxford / 2013 / 10

For the approaching ship, ${\beta }_s=0.85$ and ${\gamma }_s=(1-0.85^2{)}^{-1/2}$. Its station-frame length contraction gives $L_s=100/{\gamma }_s\mathrm{m}$. Adding the two station-frame clearances givesThis is the berth's proper length, since the berth is at rest in the station. It is a simultaneous station-frame allocation, not a ship-frame guarantee of a safe docking maneuver.

The berth moves at $0.85c$ in the ship's initial rest frame, so it too is subject to length contraction:The ship remains $100\mathrm{m}$ long in this frame. The two length comparisons use different simultaneous endpoint events; this is the same relativity of simultaneity that resolves the ladder paradox. Physical docking also requires deceleration, whose later effects are outside these initial inertial measurements.

Interpret the one-light-day separation $D$ as measured in the station frame at capsule launch, and retain the ship's initial inertial rest frame until reception at the station. The phrase does not explicitly name a frame; an alternative is given below. The ship-frame simultaneous separation is $D^{\prime }=D/{\gamma }_s$. The capsule moves at $0.9c$ towards a station approaching at $0.85c$, so their separation decreases at $1.75c$ in this frame. This is a separation rate, not either object's speed. Hence the elapsed ship-frame coordinate time isIt is the interval assigned using synchronized clocks in the ship frame to the launch and remote arrival. A single onboard clock does not directly mark the remote arrival event; observing its arrival signal would add a light-travel delay. If the stated one-light-day separation were instead measured in the ship frame, the answer would be $\Delta t_s=(1/1.75)\mathrm{day}=0.571429\mathrm{day}$.

Launch and arrival occur on the capsule's own worldline, so its clock measures the proper time. In the ship frame its Lorentz factor is ${\gamma }_{c/s}=(1-0.9^2{)}^{-1/2}$; thereforeThis uses the station-frame distance convention stated in i. With a one-light-day distance in the ship frame instead, it becomes $\Delta {\tau }_c=\sqrt{0.19}/1.75\mathrm{day}\simeq 0.249080\mathrm{day}$. The uncertainty is a distance-frame convention in the wording, not a missing figure or an unanswered mathematical part.

The velocity-addition formula combines the ship's station-frame velocity and the capsule's ship-frame velocity:For the distance convention in i, station-frame arrival takes $D/v_c=(353/350)\mathrm{day}=1.008571\mathrm{day}$. Transforming its two endpoint events back to the ship frame gives ${\gamma }_s[D/v_c-0.85D/c]$, agreeing with i. The capsule remains below the speed of light in either frame.

The capsule continues inertially at $v_c/c=350/353$ while the ship reduces its station-frame speed to $0.7c$. Transform the ship's new velocity into the capsule frame:Thus its speed is $0.952778c$, and the negative sign means it recedes behind the capsule along the chosen axis. The same transformation compares the subsequent local inertial velocities; the details of the intervening deceleration are not needed.