Orbital Debris Collision
The International Space Station has hundreds of shields made of aluminum and bulletproof composite materials that are intended to offer protection against impact with debris present in low Earth orbit (Figure 3.5).With sufficient advance warning, the station’s orbit can even be adjusted slightly to avoid close approaches of larger objects. Over 13,000 pieces of debris have been identified by the United States Space Command, including paint chips, spent booster casings, and even an astronaut’s glove. (a) Calculate the kinetic energy U_{k} =\frac{1}{2} mv^{2} of an m = 1 g particle of debris traveling at v = 8 km/s, which is a typical velocity in low Earth orbit. (b) How fast would a 0.31-lb baseball have to be thrown to have the same kinetic energy?
Approach
We first convert the debris particle’s mass and velocity to the dimensionally consistent units of kg and m/s, respectively, using the definition of the “kilo” prefix (Table 3.3).
Table 3.3 Order-of-Magnitude Prefixes in the SI
| Name symbol Multiplicative Factor | ||
| tera | T | 1,000,000,000,000 = 10^{12} |
| giga | G | 1,000,000,000 = 10^9 |
| mega | M | 1,000,000 = 10^6 |
| kilo | k | 1000 = 10^3 |
| hecto | h | 100 = 10^2 |
| deca | da | 10 = 10^1 |
| deci | d | 0.1 = 10^{-1} |
| centi | c | 0.01 = 10^{-2} |
| milli | m | 0.001 = 10^{-3} |
| micro | μ | 0.000,001 = 10^{-6} |
| nano | n | 0.000,000,001 = 10^{-9} |
| pico | p | 0.000,000,000,001 = 10^{-12} |
The conventional SI unit for energy in Table 3.2
Table 3.2 Certain Derived Units in the SI
| Quantity | SI Derived Unit | Abbreviation | Definition |
| Length | micrometer or micron | \mu m | 1\mu m = 10^{-6} m |
| Volume | liter | L | 1 L = 0.001 m^{3} |
| Force | newton | N | 1 N = 1 (kg . m)/s^{2} |
| Torque, or moment of a force | newton-meter | N . m | — |
| Pressure or stress | pascal | Pa | 1 Pa = 1 N/m^{2} |
| Energy, work, or heat | joule | J | 1 J = 1 N . m |
| Power | watt | W | 1 W = 1 J/s |
| Temperature | degree Celsius | °C | °C = K – 273.15 |
| Although a change in temperature of 1 Kelvin equals a change of 1 degree Celsius, numerical values are converted using the formula. | |||
is the joule, defined as 1 N. m. In part (b), we will convert the kinetic energy to the USCS using the factor 1 J = 0.7376 ft . lb from Table 3.6.
Table 3.6 Conversion Factors between Certain Quantities in the USCS and SI
| Quantity | Conversion |
| Length | 1 in. = 25.4 mm |
| 1 in. = 0.0254 m | |
| 1 ft = 0.3048 m | |
| 1 mi = 1.609 km | |
| 1 mm = 3.9370 ×10^{-2} in. | |
| 1 m = 39.37 in. | |
| 1 m = 3.2808 ft | |
| 1 km = 0.6214 mi | |
| Area | 1 in^2 = 645.16 mm^2 |
| 1 ft^2 = 9.2903 × 10^{-2} m^2 | |
| 1 mm^2 = 1.5500 × 10^{-3 }in^2 | |
| 1 m^2 = 10.7639 ft^2 | |
| Volume | 1 ft^3 = 2.832 × 10^{-2} m^3 |
| 1 ft^3 = 28.32 L | |
| 1 gal = 3.7854 × 10^{-3} m^3 | |
| 1 gal = 3.7854 L | |
| 1 m^3 = 35.32 ft^3 | |
| 1 L = 3.532 × 10^{-2} ft^3 | |
| 1 m^3 = 264.2 gal | |
| 1 L = 0.2642 gal | |
| Mass | 1 slug = 14.5939 kg |
| 1 lbm = 0.45359 kg | |
| 1 kg = 6.8522 × 10^{-2} slugs | |
| 1 kg = 2.2046 lbm | |
| Force | 1 lb = 4.4482 N |
| 1 N = 0.22481 lb | |
| Pressure or stress | 1 psi = 6895 Pa |
| 1 psi = 6.895 kPa | |
| 1 Pa = 1.450 × 10^{-4 }psi | |
| 1 kPa = 0.1450 psi | |
| Work, energy, or heat | 1 ft · lb = 1.356 J |
| 1 Btu = 1055 J | |
| 1 J = 0.7376 ft · lb | |
| 1 J = 9.478 × 10^{-4} Btu | |
| Power | 1 (ft · lb)/s = 1.356 W |
| 1 hp = 0.7457 kW | |
| 1 W = 0.7376 (ft · lb)/s | |
| 1 kW = 1.341 hp |
Since the baseball’s weight is specified in the problem statement, we will make an intermediate calculation for its mass.
