Question

The space shuttle releases a satellite into a circular orbit 630 km above the Earth.

Answer 1

Answer:

7,539 m/s

Explanation:

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

• $\displaystyle g=\frac{GM}{r^2}$

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:

• $\displaystyle g=\frac{(6.67\cdot 10^-^1^1 \ Nm^2kg^-^2)(5.972\cdot 10^2^4 \ kg)}{[(6.3781\cdot 10^6 \ m)+(630000 \ m)]^2}$

Remove units to make the equation easier to read.

• $\displaystyle g=\frac{(6.67\cdot 10^-^1^1 )(5.972\cdot 10^2^4 )}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Multiply the numerator out.

• $\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Add the terms in the denominator.

• $\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(7008100)]^2}$

Simplify this equation.

• $\displaystyle g=8.11045189 \ \frac{m}{s^2}$

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.

• $\displaystyle a = \frac{v^2}{r}$

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.

• $\displaystyle 8.11045189 \ \frac{m}{s^2} = \frac{v^2}{7008100 \ m}$  

Remove units to make the equation easier to read.

• $\displaystyle 8.11045189 = \frac{v^2}{7008100}$

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.