The space shuttle releases a satellite into a circular orbit 630 km above the Earth. How fast must the shuttle be moving (relative to Earth's center) when the release occurs?

7,539 m/s

Explanation:

Let's use this equation to find the gravitational acceleration of this space shuttle:

We know that G is the gravitational constant: 6.67 * 10^(-11) Nm²/kg².

M is the mass of the planet, which is Earth in this case: 5.972 * 10^24 kg.

r is the distance from the center of Earth to the space shuttle: radius of Earth (6.3781 * 10^6 m) + distance above the Earth (630 km → 630,000 m).

Plug these values into the equation:

Remove units to make the equation easier to read.

Multiply the numerator out.

Add the terms in the denominator.

Simplify this equation.

The acceleration due to gravity g = 8.11045189 m/s². Now we use the equation for acceleration for an object in circular motion which contains v and r.

a = g, v is the velocity that the space shuttle should be moving (what we are trying to solve for), and r is the radius we had in the previous equation when solving for g.

Plug these values into the equation and solve for v.

Remove units to make the equation easier to read.

Multiply both sides by 7,008,100.

Take the square root of both sides.

The shuttle should be moving at a velocity of about 7,539 m/s when it is released into the circular orbit above Earth.

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