The space shuttle releases a satellite into a circular orbit 700 km above the Earth. How fast must the shuttle be moving (relative to Earth) when the release occurs?
In order to calculate the speed at which the space shuttle must be moving when it releases the satellite, we need to determine the orbital velocity of the satellite.
The orbital velocity (v) of an object can be calculated using the following equation:
v = √(GM/R)
where;
G = 6.67 × 10^(-11) N·m²·kg^(-2) (the gravitational constant)
M = 5.97 × 10^(24) kg (the mass of Earth)
R = the distance between the object and the center of Earth
We are given that the satellite is released 700 km above the Earth's surface. Since the radius of the Earth is approximately 6,371 km, the distance between the satellite and the center of the Earth (R) can be calculated as:
R = (6,371 km + 700 km) × 1,000 m/km
R = 7,071,000 m
Now, we can calculate the orbital velocity (v) using the equation:
v = √(6.67 × 10^(-11) N·m²·kg^(-2) × 5.97 × 10^(24) kg / 7,071,000 m)
v ≈ 7,448 m/s
Therefore, the space shuttle must be moving at a speed of approximately 7,448 m/s (relative to Earth) when it releases the satellite.
To determine the required speed of the space shuttle when it releases the satellite into a circular orbit, we can rely on the concept of centripetal force. The centripetal force is provided by the gravitational force between the Earth and the satellite. The formula for centripetal force is given by:
Fc = mv^2 / r
where Fc is the centripetal force, m is the mass of the satellite, v is the velocity of the satellite, and r is the radius of the circular orbit.
In this case, we need to rearrange the formula to solve for v:
v = sqrt(Fc * r / m)
Now, we can substitute the values:
Gravitational force (Fc) = G * Me * Ms / r^2
where G is the gravitational constant (6.67430 × 10^-11 N(m/kg)^2), Me is the mass of the Earth (5.972 × 10^24 kg), and Ms is the mass of the satellite.
Plugging in the values, we get:
v = sqrt((G * Me * Ms / r^2) * r / m)
The mass of the satellite cancels out, so we are left with:
v = sqrt(G * Me / r)
Substituting the known values:
v = sqrt((6.67430 × 10^-11 N(m/kg)^2) * (5.972 × 10^24 kg) / (700,000 m))
By calculating the equation, we find that the required speed for the space shuttle relative to Earth when releasing the satellite into a circular orbit 700 km above the Earth is approximately 7,751 meters per second (m/s).
To determine the speed at which the space shuttle must be moving (relative to Earth) when the satellite is released into a circular orbit 700 km above the Earth, we can make use of the concept of orbital velocity.
The orbital velocity is the minimum speed required for an object to maintain a stable orbit around the Earth. It depends on the distance from the center of the Earth to the object.
To calculate the orbital velocity, we can make use of the following formula:
Orbital velocity (V) = √(GM / r)
where:
- G represents the gravitational constant (6.67430 × 10^-11 m^3 kg^-1 s^-2)
- M represents the mass of the Earth (5.972 × 10^24 kg)
- r represents the distance between the center of the Earth and the object (in this case, the altitude of the satellite + radius of the Earth)
In this case, the altitude of the satellite is given as 700 km above the Earth's surface, so we need to calculate the distance from the center of the Earth to the satellite.
The radius of the Earth is approximately 6,371 km. Adding the altitude of the satellite (700 km) gives us the distance from the center of the Earth to the satellite:
r = 6,371 km + 700 km = 7,071 km = 7,071,000 meters
Now, we can substitute the values into the formula:
V = √((6.67430 × 10^-11 m^3 kg^-1 s^-2) * (5.972 × 10^24 kg) / 7,071,000 m)
By evaluating the expression, we will get the orbital velocity for the satellite.
After finding the orbital velocity, we can add it to the rotational velocity of the Earth at that location to obtain the speed at which the space shuttle must be moving relative to the Earth when the release occurs.