# Calculate a space shuttle territory 250 km above Earth's surface. Earth's mass is 6.0 10 24 k...

## Question:

Calculate a space shuttle territory 250 km above Earth's surface. Earth's mass is {eq}6.0 \times 10^{24}\;kg {/eq} and its radius is{eq}6.38 \times 10^6\; m {/eq} (6380km).

## Velocity:

When the object cover the distance in a unit time in a specific direction is known as velocity. The meter per second is a measurable unit to define velocity. It used to analyze the motion of the object,

## Answer and Explanation:

Given Data:

• The mass of earth is: {eq}{M_o} = 6.24 \times {10^{24}}\;{\rm{kg}} {/eq}
• The radius of earth is: {eq}{R_o} = 6.38 \times {10^6}\;{\rm{m}} {/eq}
• The height of space shuttle territory is: {eq}y = 250\;{\rm{km}} = 250000\;{\rm{m}} {/eq}

The expression for radius of space shuttle territory from the earth is

{eq}{R_s} = {R_o} + y {/eq}

The expression for acceleration due to gravity to height {eq}y = 250\;{\rm{km}} = 250000\;{\rm{m}} {/eq} from earth is

{eq}{g_s} = g\left( {1 - \dfrac{{2y}}{{{R_o}}}} \right) {/eq}

Substitute the value and solve the above expression

{eq}\begin{align*} {g_s} &= 9.81\left( {1 - \dfrac{{2\left( {250000} \right)}}{{6.38 \times {{10}^6}}}} \right)\\ &= 9.81\left( {1 - 0.0783} \right)\\ &= 9.041\;{\rm{m/}}{{\rm{s}}^2} \end{align*} {/eq}

The space shuttle travel in rotation motion and it experience centripetal acceleration that must balance by the acceleration due to gravity.

{eq}{g_s} = \dfrac{{v_s^2}}{{{R_s}}} {/eq}

Here {eq}{v_s} {/eq} is velocity of space shuttle

Substitute the value and solve the above expression

{eq}\begin{align*} 9.041 &= \dfrac{{v_s^2}}{{{R_o} + y}}\\ {v_s} &= \sqrt {9.041\left( {6.38 \times {{10}^6} + 250000} \right)} \\ &= 7.742 \times {10^3}\;{\rm{m/s}} \end{align*} {/eq}

The expression for time to cover space shuttle territory is

{eq}T = \dfrac{{2\pi {R_s}}}{{{v_s}}} {/eq}

Substitute the value and solve the above expression

{eq}\begin{align*} T &= \dfrac{{2\pi \left( {{R_o} + y} \right)}}{{7.742 \times {{10}^3}\;{\rm{m/s}}}}\\ &= \dfrac{{2\pi \left( {6.38 \times {{10}^6} + 250000} \right)}}{{7.742 \times {{10}^3}\;{\rm{m/s}}}}\\ &= 5380.717\;{\rm{s}}\left( {\dfrac{{1\;{\rm{h}}}}{{3600\;{\rm{s}}}}} \right)\\ &= 1.494\;{\rm{h}}\\ &\approx 1.5\;{\rm{h}} \end{align*} {/eq}

Thus the time require to complete space shuttle territory is {eq}1.5\;{\rm{h}} {/eq} with {eq}7.742 \times {10^3}\;{\rm{m/s}} {/eq}.