Answer:
The length contraction is given by
In this formula, L is the length of the object as seen by an observer whose
speed relative to the object is v, and 
is its ``proper length",
i.e. the length of the
object as measured in its rest frame, i.e. as measured by an observer who is at
rest with respect to the object. We are looking for that 
for which
Solving the relation between 
and 
for
, one obtains:
In our case this gives
The speed necessary is therefore = 
(You can read off this number
also
from Fig.11.6 of the textbook).
At this speed, the United States appear (5000/2 km) = 2500 km wide.
Answer:
Using the same relations as in the previous exercise, we find that the Lorentz factor
Therefore a meter stick will appear 0.8 m long. Since Velma flies
lengthwise over
the swimming pool, only its length will appear to be contracted
(to 
m = 16 m, while its width will appear unchanged.
(You can also read off the factor 0.8 from Fig. 11.2).
Answer:
According to Galilean relativity, the velocities add linearly, i.e. the total speed relative to Mort
.According to Einstein's relativity, the velocities add according to
In our case this gives
The speeds involved are too large for Galilean relativity to be applicable. Einstein's generalization needs to be used.
Answer:
The Lorentz factor
for this speed is:
To Mort, the mass of Velma's spaceship appears bigger by a factor of 
,
while its length (in the direction of motion) appears contracted by a factor
. Therfore, according to Mort, Velma's spaceship has a mass of
16666.7 kg, and a length of 60 m. (You can also get these numbers from
Fig. 11.4).
Answer:
As seen from you, Mort is moving past you at nearly light speed. Therefore his mass appears increased by a large factor, his heartbeat has slowed down, and his size (in the direction of motion) has shrunk by the same factor.
Answer:
The mass loss in fission is about 1%, i.e. in the fission process of 1 kg uranium, 0.01 kg is converted into thermal energy. The corresponding energy is
To heat 1 kg water from freezing to boiling point requires
.
With 
J, one can heat
i.e. about 2 billion kg of water.
? What if you were not accelerating at all?
Answer:
If the rocketship were accelerating at 2 g, the ball would fall toward the rear (which you would perceive as ``down") with an acceleration of 2 g. If the acceleration of the rocketship were
,
the acceleration of the ball ``downwards" (i.e. towards the rear) would be
. If the rocketship were not accelerating at all, the dropped ball would
stay suspended where you released it.
Answer:
In the (rather good) approximation that the Earth is a sphere, the longitudinal lines on Earth, also called ``meridians", are circles. They are ``great circles" (i.e the largest circles one can draw on the surface of a sphere, with center = the center of the sphere, and radius = radius of the sphere). Such great circles are ``geodesic lines" (i.e the shortest connection between two points) in the two-dimensional space formed by the surface of the sphere.
The latitudinal lines are also circles, but not great circles. They do not meet (intersect), but they are not geodesic (``straightest") lines, apart from one exception, the equator (= the 0 latitude line).
ANswer:
A circle is a curved one-dimensional space. A straight line (infinitely long) is a flat one-dimensional space.