Using the inverse-square gravitational law, calculate the work required to lift a 1000-kg satellite from the earth's surface to an altitude of 100 mi and inject it into orbit at 17,500 mph. How long would a 300-hp engine have to be run to produce this much energy
To calculate the work required to lift the satellite, we need to consider both the gravitational potential energy and the kinetic energy of the satellite.
Step 1: Calculate the gravitational potential energy:
The gravitational potential energy can be calculated using the formula:
PE = mgh
where PE is the potential energy, m is the mass of the satellite, g is the acceleration due to gravity, and h is the height.
Given:
Mass of satellite (m): 1000 kg
Acceleration due to gravity (g): 9.8 m/s^2
Height (h): 100 miles = 160,934 meters
PE = (1000 kg) x (9.8 m/s^2) x (160,934 m)
PE = 1.568 x 10^10 Joules
Step 2: Calculate the kinetic energy:
The kinetic energy can be calculated using the formula:
KE = (1/2)mv^2
where KE is the kinetic energy, m is the mass of the satellite, and v is the velocity.
Given:
Mass of satellite (m): 1000 kg
Velocity (v): 17,500 mph = 27778 m/s
KE = (1/2) x (1000 kg) x (27778 m/s)^2
KE = 3.854 x 10^11 Joules
Step 3: Calculate the total work:
The total work required to lift the satellite is the sum of the gravitational potential energy and the kinetic energy.
Total Work = PE + KE
Total Work = 1.568 x 10^10 Joules + 3.854 x 10^11 Joules
Total Work = 4.011 x 10^11 Joules
Step 4: Calculate the time required by a 300-hp engine:
To calculate the time required, we need to convert the work into energy and then divide it by the power of the engine.
Given:
Power (P) of 300-hp engine: 300 hp = 223,710 Watts
Energy = Power x Time
Time = Energy / Power
Energy = 4.011 x 10^11 Joules
Power = 223,710 Watts
Time = (4.011 x 10^11 Joules) / (223,710 Watts)
Time = 1.793 x 10^6 seconds
Therefore, a 300-hp engine would have to be run for approximately 1.793 x 10^6 seconds to produce the required energy.