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?

700,000 = 7 * 10^5 meters above so

r = 7*10^5 + 6.38*10^6 = 70.8*10^5 = 7.08*10^6

m v^2/r = G m M/r^2

v^2 = G M/r

v^2 = 6.67*10-11 * 6*10^24 / 7.08*10^6

v^2 = 5.65 * 10^7 = 56.5*10^6

v = 7.5*10^3 = 7500 m/s

To find the speed at which the space shuttle must be moving to release the satellite into a circular orbit 700 km above the Earth, we can use the concept of centripetal force.

The centripetal force required to keep an object in circular motion is provided by the gravitational force of Earth. The gravitational force is given by the formula:

Fg = (G * m1 * m2) / r^2

Where Fg is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers.

In this case, we have the mass of the satellite (m1) and the mass of the Earth (m2). The force required to keep the satellite in circular motion is equal to the gravitational force:

Fc = Fg

The centripetal force is given by the formula:

Fc = (m * v^2) / r

Where m is the mass of the satellite, v is the velocity of the satellite in circular orbit, and r is the distance from the center of Earth to the satellite.

Setting Fc equal to Fg, we have:

(m * v^2) / r = (G * m1 * m2) / r^2

We can cancel out the masses of the satellite and Earth:

v^2 / r = (G * m2) / r^2

Rearranging the equation to find v:

v^2 = (G * m2) / r

v = sqrt((G * m2) / r)

Now we can plug in the values:

G = 6.674 * 10^-11 N m^2 / kg^2 (gravitational constant)
m2 = 5.972 * 10^24 kg (mass of Earth)
r = 700,000 m (distance from Earth's center to the satellite)

Calculating the velocity:

v = sqrt((6.674 * 10^-11 N m^2 / kg^2 * 5.972 * 10^24 kg) / 700,000 m)

Using a calculator, we find:

v ≈ 7,902 m/s

Therefore, the shuttle must be moving at a speed of approximately 7,902 m/s relative to Earth when the release occurs.

To determine the speed at which the space shuttle must be moving when it releases the satellite into a circular orbit 700 km above the Earth, you need to use the concept of centripetal force.

The centripetal force required to keep an object in circular motion is given by the equation:

F = (mv^2) / r

where F is the centripetal force, m is the mass of the object, v is the velocity, and r is the radius of the circular orbit.

In this case, the satellite is in orbit around the Earth, so the centripetal force is provided by the gravitational force between the Earth and the satellite. The gravitational force is given by the equation:

F = (G * M * m) / r^2

where G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the radius of the orbit.

Since the satellite is in a circular orbit, both of these forces must be equal. Therefore, we can set the equations for gravitational force and centripetal force equal to each other:

(G * M * m) / r^2 = (mv^2) / r

We can now cancel out the mass (m) from both sides of the equation:

(G * M) / r = v^2

To solve for v, we take the square root of both sides:

v = √((G * M) / r)

Plugging in the given values:

G = 6.674 × 10^-11 m^3/(kg*s^2) (gravitational constant)
M = 5.972 × 10^24 kg (mass of the Earth)
r = 700,000 m (radius of the orbit)

v = √((6.674 × 10^-11 m^3/(kg*s^2) * 5.972 × 10^24 kg) / 700,000 m)

Calculating this expression, you will find that the speed at which the shuttle must be moving (relative to Earth) when the satellite is released into a circular orbit 700 km above the Earth is approximately 7,660 meters per second.