This **article aims to determine the speed of the satellite**Â $B$. This article uses the **concept of speed. **In physics, **uniform circular motion**Â describes **motion of a body**Â traversing a **circular path**Â at a constant speed.

Since the body describes **circular motion**, its **distance from the axis of rotation remains constant throughout**. Although a**Â body’s speed is constant**, its speed is not constant. **V****elocity, a vector quantity**, depends on both the body’s **speed**Â and**Â direction.**

**Expert Answer**

**Given **

\[OA = r \]

\[OB Â = Â 2r \]

\[m_{A} = m \]

\[m_{B} = 2m\]

\[v_{A} = 1000 \dfrac {m}{s} \]

The**Â speed of satellite**Â $ B $ is given as:

\[m_{p} = mass\: of \:the \:planet \]

\[\dfrac {m_{A} . g_{A} }{m_{B}.g_{B}} = \dfrac{K. \dfrac{m_{A}.m_{P}}{r ^ {2}}}{\dfrac{m_{B}.m_{P}}{4r ^ {2}}}\]

**Equation**Â $ 1 $ is given as:

\[\dfrac{g_{A}}{g_{B}} = 4. \dfrac{m_{A}} {m_{B}} \]

\[\dfrac{m_{A} . g_{A} }{m_{B}.g_{B}} = \dfrac{\dfrac{m_{A}.v_{A} ^ {2}}{r ^ {2}}}{\dfrac{m_{B}.v_{B} ^ {2}}{4r ^ {2}}}\]

**Equation**Â $ 2 $ is given as:

\[\dfrac{g_{A}}{g_{B}} = 4. \dfrac{v_{A} ^ {2}}{v_{B} ^ {2}}\]

By **equalizing both equations**Â $ 1 $ and $ 2 $:

\[\dfrac{m_{A}}{m_{B}} = \dfrac{v_{A} ^ {2}}{v_{B} ^ {2}}\]

\[m_{A} = m\: ; \: m_{B} = 2m;\]

\[\dfrac{m}{2m} = \dfrac{v_{A}^{2}}{v_{B} ^ {2}}\]

\[ v_{B} Â = v_{A}.\sqrt {2} \]

\[ v_{A} Â = Â 10000 \dfrac {m} {s} \]

\[ v_{B} Â = Â 10000 \sqrt 2 \dfrac {m} {s} \]

**Numerical Result**

The speed of satellite $ B $ is $10000 \sqrt 2 \dfrac {m}{s}$.

**Example**

Satellite $ A $ orbits the planet at a speed of $20,000 \dfrac{m}{s}$. Satellite $ B $ is twice as massive as satellite $ A $ and orbits at twice the distance from the center of the planet. What is the speed of satellite $ B $? Assume that both orbits are circular.

**Solution**

**Given **

\[OA = r \]

\[OB = 2r \]

\[m_{A} = m \]

\[m_{B} = 2m \]

\[v_{A} = 2000 \dfrac {m} {s} \]

The**Â speed of satellite**Â $B$ is given as:

\[m_{p} = Â mass\: of \:the \:planet \]

\[\dfrac { m_{A} . g_{A} } {m_{B} . g_{B} } = \dfrac{K. \dfrac {m_{A} . m_{P} }{r ^ {2}} }{ \dfrac{m_{B} . m_{P} }{4r ^ {2} } } \]

**Equation**Â $1$ is given as:

\[\dfrac {g_{A} } {g_{B} } = 4 . \dfrac{m_{A}} {m_{B}} \]

\[\dfrac {m_{A} . g_{A} } {m_{B}.g_{B} } = \dfrac {\dfrac{m_{A}.v_{A} ^ {2}}{r ^ {2}} }{ \dfrac { m_{B} . v_{B} ^ {2} }{ 4r ^ {2}} }\]

**Equation**Â $2$ is given as:

\[\dfrac {g_{A} }{ g_{B} } = 4. \dfrac {v_{A} ^ {2} }{v_{B} ^ {2} }\]

By **equalizing both equations**Â $ 1 $ and $ 2 $:

\[\dfrac {m_{A} }{m_{B} } = \dfrac {v_{A} ^ {2} }{v_{B} ^ {2} } \]

\[m_{A} = m\: ; \: m_{B} = 2m;\]

\[\dfrac { m }{ 2m } = \dfrac { v_{A} ^ {2} }{v_{B} ^ {2} } \]

\[v_{B} = v_{A} . \sqrt {2} \]

\[v_{A} = 20000 \dfrac {m} {s} \]

\[v_{B} = 20000 \sqrt 2 \dfrac {m} {s} \]

The **speed of satellite**Â $B$ is $20000 \sqrt 2 \dfrac {m} {s} $.