The space shuttle releases a satellite into a circular orbit 550 km above the Earth. How fast must the shuttle be moving (relative to Earth) when the release occurs?

The earth's radius is Re = 6378 km, so the radius of the orbit is

R = Re + 550 = 6928 km

For a circular orbit, the centripetal force is the gravatational attraction force, so
V^2/R = G M/R^2

G is the universal constant of gravity and M is the mass of the earth.

Solve for V.

V = sqrt(GM/R) = sqrt[(GM/Re^2)*Re^2/R]
= sqrt(g Re)*sqrt(Re/R)
= 0.956*sqrt(g Re)

Why did the space shuttle release a satellite? So it can have some space on its own, of course! Anyway, to answer your question, the shuttle must be moving with a velocity of about 7.9 kilometers per second (or about 28, 400 kilometers per hour) relative to the Earth when it releases the satellite. That's one speedy shuttle! So fast, it could probably give Usain Bolt a run for his money!

To determine the speed of the space shuttle when it releases the satellite into a circular orbit 550 km above the Earth, we can use the concept of orbital velocity.

Orbital velocity is the minimum velocity required for an object to maintain a stable orbit around the Earth. It can be calculated using the formula:

v = sqrt(G * M_e / r)

Where:
v = orbital velocity (in m/s)
G = gravitational constant (approximately 6.67430 × 10^-11 N(m/kg)^2)
M_e = mass of the Earth (approximately 5.972 × 10^24 kg)
r = radius of the orbit (in meters)

First, we need to convert the given distance of 550 km to meters:
550 km = 550,000 m

Plugging the values into the formula, we get:

v = sqrt((6.67430 × 10^-11 N(m/kg)^2) * (5.972 × 10^24 kg) / (550,000 m + 6,371,000 m))

Simplifying further:

v = sqrt(39.47842 m^3/kg/s^2)

Taking the square root:

v ≈ 6,284 m/s

Therefore, the shuttle must be moving at approximately 6,284 m/s (relative to the Earth) when it releases the satellite to maintain a circular orbit 550 km above the Earth.

To find the speed of the space shuttle when it releases the satellite into a circular orbit, we can use the concept of centripetal force.

When an object is in a circular orbit, it experiences a centripetal force that keeps it moving in a circle. This force is provided by the gravitational force between the object and Earth.

The centripetal force can be calculated using the equation:

F = (mv^2)/r,

where F is the centripetal force, m is the mass of the satellite, v is its velocity, and r is the distance between the satellite and the center of the Earth.

In this case, the centripetal force is provided by the gravitational force, which can be expressed as:

F = (GmM)/(r^2),

where G is the universal gravitational constant and M is the mass of the Earth.

Since these two forces are the same, we can set them equal to each other and solve for v:

(mv^2)/r = (GmM)/(r^2).

Canceling out the mass of the satellite (m) and simplifying, we get:

v^2 = (GM)/r.

Now, we can plug in the known values:

G = 6.67 × 10^-11 N(m/kg)^2 (gravitational constant).
M = 5.97 × 10^24 kg (mass of the Earth).
r = 550 km = 550,000 m (distance from the satellite to the center of the Earth).

Plugging in these values, we find:

v^2 = ((6.67 × 10^-11 N(m/kg)^2) × (5.97 × 10^24 kg))/(550,000 m).

Calculating this expression, we can find v^2, the square of the velocity of the shuttle when it releases the satellite.

Finally, taking the square root of v^2 will give us the required speed of the shuttle.

By following this calculation, you should be able to determine the velocity of the shuttle when it releases the satellite into a circular orbit.