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

The satellite is lauched into orbit 200 km above Earth.

Note: 1 km=2000 meters.

Therefore,

$\begin{array}{l}200\text{​â¶Ä‹â¶Ä‰â¶Ä‰k³¾}=200×1000\\ =200000\text{ â¶Ä‰â¶Ä‰meters}\end{array}$

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,

$\begin{array}{c}r=200000+6380000\\ =6580000\text{ â¶Ä³¾±ð³Ù±ð°ù²õ}\end{array}$

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

The velocity needed for the artificial satellite to achieve orbit around the earth is called as 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$5.97×{10}^{24}$ kilograms.

The value of $G$is $6.67×{10}^{−11}\text{ â¶Ä‰}\frac{{\text{Nm}}^{\text{2}}}{{\text{kg}}^{\text{2}}}$.

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.

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

Hence, the orbital velocity of the satellite’s is $0.78×{10}^4\text{ â¶Ä‰â¶Ä‰}m/s$.