simple mathematics language
. An elementary particle called a pi meson (or pion for short) has an average lifetime of 2.6 × 10−8 when at rest. If a pion moves with a speed of 0.89crelative to Earth, find (a) the average lifetime of the pion as measured by an observer on Earth and (b) the average distance traveled by the pion as measured by the same observer. (c) How far would the pion have traveled relative to Earth if relativistic time dilation did not occur?
a)To determine the average lifetime of the pion measured by an observer on Earth, the expression for time dilation is used:
=
(Walker pg. 1021)
Where represents the dilated time measured by the observer relative to the event; the lifetime of the pion. The proper time is represented by
, and is “the amount of time separating two events which occur at the same location” (Walker pg. 1022), the proper time is given as the average lifetime of the pion when at rest, 2.6 × 10−8 sec. The pion moves at a speed of .89c relative to the location of the observer and pion, Earth. The speed of light in a vacuum is represented by c and is equal to 3.00 x 108 m/s (Walker pg. 884). The pion moves at a speed 89% that of light, the speed of light is given a value of 1, as it is constant. Filling in the numerical values:
=
=
=
= 5.70225 x 10-8sec
The average lifetime of the pion measured by an observer on Earth is 5.7 x 10-8sec.
b) The average distance traveled by the pion measured by the same observer is given by the velocity:
v =
Solving for the distance:
v() =
(
)
d = v()
The velocity, v, is calculated by multiplying the speed (0.89c) and the speed of light (3.00 x 108 m/s):
v = (0.89c) (3.00 x 108 m/s)
v = 2.67 x 108m/s
Filling in the values for the velocity (2.67 x 108m/s) and the dilated time (5.7 x 10-8sec):
d = (2.67 x 108m/s) (5.7 x 10-8sec)
d = 15.219 m
The average distance traveled by the pion measured by the observer is 15 meters.
c)In the evet that relativistic time dilation did not occur, that is the change in time equals the proper time, the distance the pion traveled relative to Earth is:
d = v()
d = (2.67 x 108m/s) )
d = 6.942m
If relativistic time dilation did not occur the pion traveled 6.9 meters.
24. A rectangular painting is W = 117 cm wide and H = 76.2 cm high, as indicated in the figure. At what speed, υ, must the painting move parallel to its width if it is to appear to be square?
As objects move, the length of the object will contract or shorten in the direction of their motion relative to the object’s length when at rest. To determine the speed the painting must travel in order to achieve a change in shape, apply the expression which defines the contracted length L in terms of the proper length Lo and the velocity which the painting moves:
L = Lo (Walker pg. 1026)
Solving for the velocity:
L2 = Lo2(1 – )
Dividing both sides by Lo2:
= (1 –
)
= (1 –
)
v = c (Walker pg. 1027)
The proper length, is the width of the rectangular painting as 117 cm, the contracted length,
of the painting is 76.2 cm. Filling in the numerical values for the proper length, the contracted length, and the speed of light in a vacuum, expressed as 1, as it is constant:
v = 1c
v = 0.758c
The painting must move at a speed of 0.758 c.
34. A spaceship moving toward Earth with a speed of 0.85c launches a probe in the forward direction with a speed of 0.15c relative to the ship. Find the speed of the probe relative to Earth.
The speed of the probe relative to Earth is determined using the expression for relativistic addition of velocities:
v23 = (Walker pg. 1029)
Where v23 is the speed of the probe relative to Earth, v21 is the speed of the probe relative to the ship and is given as 0.15c, v13 is the speed of the ship relative to the Earth and is given as 0.85c. Filling in the numerical values for v21 and v13 ( where the speed of light in a vacuum is 1c):
v23 =
v23 = 0.8869c
The speed of the probe relative to the Earth is 0.89c.
45. An object has a relativistic momentum that is 8.50 times greater than its classical momentum. What is its speed?
The classic definition of momentum is not valid for speeds approaching the speed of light. The “non-relativistic relationship p = mv must be modified for speeds comparable to c” (Walker pg. 1031). The valid expression for relativistic momentum is:
p = (Walker pg. 1031)
The speed of the object is obtained using the expression for relativistic momentum:
p =
The relativistic momentum is 8.50 times greater than its classical momentum, and is expressed as:
pR = 8.50 pC
Setting the relativistic momentum equal to the classic momentum and solving for the speed, v, where the classical momentum is p = mv:
= 8.50 mv
The terms mv reduce to 1 in the numerator on the left-hand side of the expression and on the right-hand side. Squaring the reciprocal of 8.50 on the right-hand side:
1 – =
= 1-
=
v = c
v = 0.993c
The object has a speed of 0.993c.
67. Recent measurements show that the black hole at the center of the Milky Way galaxy, which is believed to coincide with the powerful radio source Sagittarius A*, is 2.6 million times more massive than the Sun; that is, 5.2 × 1036 kg. (a) What is the maximum radius of this black hole? (b) Find the acceleration of gravity at the Schwarzschild radius of this black hole, using the expression for R given in R = . (c) How does your answer to part (b) change if the mass of the black hole is doubled? Explain.
a) The maximum radius of the black hole at the center of the Milky Way galaxy is found using the Schwarzschild radius:
R = (Walker pg. 1042)
Where the mass of the black hole is represented by M, the gravitational constant is represented as G, and is 6.67 x 10-11Nm2/kg2, and the speed of light in a vacuum is represented as C and is equal to 3.00 x 108m/s. Filling in the numerical values for the gravitational constant, the mass of the black hole, 5.2 × 1036 kg, and the speed of light in a vacuum:
R =
R = 7.7075 x 109m
Converting the radius from m to km:
7.7075 x 109m x = 7.7075 x 106 km
The maximum radius of the black hole at the center of the Milky Way galaxy is 7.7 x 106km.
b)The acceleration of gravity is found by setting force in terms of mass and the acceleration of gravity F = mg equal to Newton’s law of Universal gravitation which can be applied to calculate the acceleration due to gravity on all astronomical bodies:
mg = G
The mass on each side reduces to one:
g =
Substituting the value of the Schwarzschild radius in for :
R =
g =
g =
The gravitational constant and the mass in the numerator reduce to one:
g =
Filling in the numerical values for the gravitational constant, the mass of the black hole, 5.2 × 1036 kg, and the speed of light in a vacuum:
g =
g = 5.8384 x 106m/s2
The acceleration of gravity at the Schwarzschild radius of the black hole is 5.8 x 106m/s2.
c) The expression derived for the acceleration of gravity at the Schwarzschild radius of the black hole is:
g =
This expression describes the relationship between the acceleration of gravity and the mass of the black hole as being inversely proportional to one another:
g
If the mass of the black hole is doubled, the acceleration of gravity decreases by a factor of two. This can be demonstrated by taking the given mass, 5.2 × 1036 kg, and doubling it and solving for the acceleration of gravity. The mass of the black hole when doubled equals:
M = 5.2 × 1036 kg (2) = 1.04 x 1037 kg
Setting up the acceleration of gravity as a ratio:
=
The values of the speed of light, c 4, as well as 4G in the denominator reduce to one:
=
=
Filling in the value for the original mass and the doubled mass:
=
= 2
Therefore, as the mass of the black hole is doubled, or halved, the acceleration of gravity will decrease or increase, respectively, by a factor of two.
Walker, James S. (2016). Physics. Fifth ed. San Francisco, CA: Pearson Addison-Wesley, Pearson Education, Inc.
