The Physics
Hypertextbook
Opus in profectus

# Miscellaneous Units

## Discussion

### astronomical bodies

• acceleration due to gravity on the surface of the Earth
• 1 g = 9.80665 m/s2 exactly by definition
• re or r = 6371 km
• earth mass
• me or m = 5.9736 × 1026 kg
• jupiter mass (for exoplanets)
• mj or m = 1.8986 × 1029 kg
• astronomical unit [slide]
• 1 au = 149,597,870.691 kilometers (NASA)
• solar masses
• ms or m = 1.989 × 1030 kg
• galactic year (cosmic year)
• 225 million years, but I've seen 250 million as well
• crab nebula
• "The Crab is arguably the most exhaustively studied celestial object beyond the solar system, and the apparent stability of its luminosity at all wavelengths has made it an attractive reference for calibrating observing instruments. The brightness of other objects is often quoted in "millicrabs." But the startling observation of a powerful gamma-ray flare from the Crab last September belies that vaunted stability, and it challenges the accepted theory of how charged particles are accelerated inside supernova remnants. That theory has its beginnings in Enrico Fermi's 1949 attempt to explain the origin of cosmic rays." doi:10.1063/1.3563807

### light year

1. Average speed

This problem is deceptively easy. Averaging is taught in elementary school, which makes this an elementary problem. Right?

 6.0 km/h + 5.0 km/h = 5.5 km/h 2 The wrong method of averaging

The add-and-divide method of averaging only works when averaging items of equal weight. The average age of the students in a classroom is the sum of their ages divided by the number of students only because each student is considered to have the same significance (a student, is a student, is a student,… ). In this problem, however, the two segments of the walk are significantly different. The second "half" was actually the majority of the walk. It carries more weight than the shorter first "half". Thus, the add-and-divide method won't work.

Let's return to our definition. Since speed is the rate of change of distance with time, we'll need both the distance traveled and the time it took to complete the walk. After we determine both of these numbers, the rest is easy.

Δt =
 Δs v

Δt1 =  6.0 km  = 1.0 h
6.0 km/h
Δt2 =  10 km  = 2.0 h
5.0 km/h
v =
 Δs Δt

v =  6.0 km + 10 km
1.0 h + 2.0 h
v = 5.3 km/h

Look closely at the calculations on the right side. Notice that the formula contains Δ (delta) symbols and yet I added the distances in the numerator and the times in the denominator. That's because Δ doesn't mean difference, it means change. During the walk my position didn't change from 6.0 km to 10 km, it changed first by 6.0 km and then by 10 km for a total change of 16 km.

2. Average velocity

Velocity is the rate of change of displacement with respect to time. Velocity is a vector, which means the problem should be solved graphically. Draw an arrow pointing toward the top of the page (north). Label it 6 km. Draw another arrow to the left (west) starting from the previous one (arranged head to tail). Make it slightly longer and label it 10 km. Draw a third arrow starting on the tail of the first and ending on the head of the second. Since north and west are at right angles to one another, the resultant displacement is the hypotenuse of a right triangle. Use Pythagorean theorem to find its magnitude and tangent to find its direction. r = √[(6.0 km)2 + (10 km)2] r = 11.6619… km

 tan θ = opposite = 10 km adjacent 6.0 km

θ = 59° on the west side of north

Divide displacement by time to get velocity.

v =
 Δs Δt

v =
 11.6619… km at 59° W of N 3.0 h

v = 3.9 km/h at 59° W of N

parallax second

### human constructions

• stories — typical multiunit, residential building
• 3 meters
• 10 feet (3.048 m)
• city blocks — gridded cities
• 1/20 mile (80.47 m): New York
• 1/16 mile (100.6 m): Houston, Milwaukee
• 1/8 mile (201.2 m): Chicago
• olympic swimming pools
• 50 × 25 × 2 m = 2500 m3 = 2,500,000 L
• football fields — International, American, Candian or Australian?
• 100 m
• 100 yards (91.44 m)
• cricket pitches — between the wickets
• 1 chain, 1/80 mile, 22 yards (20.12 m)
• bananas
• banana equivalent dose
• "Bananas produce antimatter, releasing one positron—the antimatter equivalent of an electron—about every 75 minutes. This occurs because bananas contain a small amount of potassium-40, a naturally occurring isotope of potassium. As potassium-40 decays, it occasionally spits out a positron in the process."