Question

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?

Answer 1

7,539 m/s

Step-by-step explanation:

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

• 
  `\displaystyle g=(GM)/(r^2)`

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:

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  `\displaystyle g=((6.67\cdot 10^-^1^1 \ Nm^2kg^-^2)(5.972\cdot 10^2^4 \ kg))/([(6.3781\cdot 10^6 \ m)+(630000 \ m)]^2)`

Remove units to make the equation easier to read.

• 
  `\displaystyle g=((6.67\cdot 10^-^1^1 )(5.972\cdot 10^2^4 ))/([(6.3781\cdot 10^6)+(630000 )]^2)`

Multiply the numerator out.

• 
  `\displaystyle g=((3.983324\cdot 10^1^4))/([(6.3781\cdot 10^6)+(630000 )]^2)`

Add the terms in the denominator.

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  `\displaystyle g=((3.983324\cdot 10^1^4))/([(7008100)]^2)`

Simplify this equation.

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  `\displaystyle g=8.11045189 \ (m)/(s^2)`

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.

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  `\displaystyle a = (v^2)/(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.

• 
  `\displaystyle 8.11045189 \ (m)/(s^2) = (v^2)/(7008100 \ m)`

Remove units to make the equation easier to read.

• 
  `\displaystyle 8.11045189 = (v^2)/(7008100)`

Multiply both sides by 7,008,100.

• 
  `56838857.89=v^2`

Take the square root of both sides.

• 
  `v=7539.154985`

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