The space shuttle releases a satellite into a circular orbit 440 km above the Earth. How fast must the shuttle be moving (relative to Earth) when the release occurs?
V^2/R = GM/R^2 for circular orbits.
Solve for V
R is the distance from the CENTER of the earth. You will need the Earth's radius, 6378 km. For R, add 440 to that and convert to meters.
G is the universal gravity constant. Get to know it.
To determine the speed of the space shuttle at the time of satellite release, we can apply the principle of conservation of energy.
The satellite will be released into a circular orbit, which means its gravitational potential energy will be equal to its kinetic energy. We can express this as:
G * M * m / r = (1/2) * m * v^2
Where:
G is the gravitational constant (approximately 6.67430 × 10^-11 N·(m/kg)^2)
M is the mass of the Earth (approximately 5.972 × 10^24 kg)
m is the mass of the satellite (which cancels out in this equation)
r is the radius of the orbit (440 km above Earth's surface = (440 + 6371) km = 6811 km = 6,811,000 m)
v is the speed of the satellite
Simplifying the equation, we can solve for v:
G * M / r = v^2
Taking the square root of both sides:
v = √(G * M / r)
Now we can plug in the values:
G = 6.67430 × 10^-11 N·(m/kg)^2
M = 5.972 × 10^24 kg
r = 6,811,000 m
v = √[(6.67430 × 10^-11 N·(m/kg)^2) * (5.972 × 10^24 kg) / (6,811,000 m)]
After evaluating this expression, we find that the speed of the space shuttle, relative to Earth, when the release occurs is approximately 7,686.7 meters per second.