Q13. - Algebra 1

Question

A satellite is launched into orbit 200 kilometers above Earth. The orbital velocity of a satellite is given by the formula $\mathit{v}\mathbf{=}\sqrt{\frac{\mathbf{G}{\mathbf{m}}_{\mathbf{E}}}{\mathbf{r}}}$. $\mathit{v}$ is velocity in meters per second, $\mathit{G}$ is a given constant, ${\mathit{m}}_{\mathbf{E}}$ is the mass of Earth, and $\mathit{r}$ is the radius of the satellite’s orbit in meters.

 

a. The radius of Earth is 6,380,000 meters. What is the radius of the satellite’s orbit in meters?

b. The mass of Earth is $\mathbf{5}\mathbf{.97}\mathbf{\times }{\mathbf{10}}^{\mathbf{24}}$ kilograms, and the constant $\mathit{G}$ is $\mathbf{6}\mathbf{.67}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{11}}\frac{{\mathbf{\text{Nm}}}^{\mathbf{\text{2}}}}{{\mathbf{\text{kg}}}^{\mathbf{\text{2}}}}$where $\mathit{N}$ is in Newtons. Use the formula to find the orbital velocity of the satellite in meters per second.

Step-by-Step Solution

Answer

a. The radius of the satellite’s orbit is 6,580,000 meters.

b.  The orbital velocity of the satellite’s is $\mathbf{0}\mathbf{.78}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathit{m}\mathbf{/}\mathit{s}$.

Part a Step 1. Explain the term ‘orbit’.

A fix repeating path that one object takes around other object (in this case artificial satellite around earth) is called orbit.

Part a Step 2. Calculate the radius of the satellite’s orbit.

The satellite is lauched into orbit 200 km above Earth.

Note: 1 km=2000 meters.

Therefore,

$\begin{array}{l}200\text{\hspace{0.17em}\hspace{0.17em}km}=200\times 1000\\ =200000\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}meters}\end{array}$

The radius of the earth is 6380000 meters.

Therefore radius of the satellite orbit$\left(r\right)$ is the total distance from the center of the earth to the orbit. That is,

$\begin{array}{c}r=200000+6380000\\ =6580000\text{\hspace{0.17em}\hspace{0.17em}meters}\end{array}$

Part a Step 3. State the conclusion.

Hence, the radius of the satellite’s orbit is 6,580,000 meters.

Part b Step 1. Explain the term ‘orbital velocity’.

The velocity needed for the artificial satellite to achieve orbit around the earth is called as orbital velocity.

Part b Step 2. Calculate the orbital velocity.

The orbital velocity of a satellite is given by the formula .

Where $v$ is velocity in meters per second, $G$ is a given constant, is the mass of Earth, and $r$ is the radius of the satellite’s orbit in meters.

The mass of the earth( $m_E$) is $5.97\times {10}^{24}$ kilograms.

The value of $G$ is $6.67\times {10}^{-11}\frac{{\text{Nm}}^{\text{2}}}{{\text{kg}}^{\text{2}}}$.

From part $\left(a\right)$ the radius of the satellite’s orbit$\left(r\right)$ is 6,580,000 meters.

Substitute these values in the formula $v=\sqrt{\frac{G{m}_E}{r}}$ and calculate the orbital velocity.

 $\begin{array}{c}v=\sqrt{\frac{G{m}_E}{r}}\\ =\sqrt{\frac{\left(6.67\times {10}^{-11}\right)\times \left(5.97\times {10}^{24}\right)}{6580000}}\\ =\sqrt{\frac{(6.67\times 5.97)\times \left({10}^{-11}\times {10}^{24}\right)}{65.8\times {10}^5}}\\ =\sqrt{\frac{39.82\times {10}^{-11+24}}{65.8\times {10}^5}}\\ =\sqrt{\frac{39.82\times {10}^{13}}{65.8\times {10}^5}}\\ =\sqrt{\left(\frac{39.82}{65.8}\right)\times \left(\frac{{10}^{13}}{{10}^5}\right)}\\ =\sqrt{\left(\frac{39.82}{65.8}\right)\times {10}^{13-5}}\\ =\sqrt{\left(\frac{39.82}{65.8}\right)\times {10}^8}\\ =\sqrt{\left(\frac{39.82}{65.8}\right)}\times \sqrt{{\left({10}^4\right)}^2}[{10}^8={\left({10}^4\right)}^2]\\ =0.78\times {10}^{4}m/s\end{array}$

Part b Step 3. State the conclusion.

Hence, the orbital velocity of the satellite’s is $0.78\times {10}^{4}m/s$.