**article aims to determine the speed of the satellite**$B$. This article uses the

**concept of speed.**In physics,

**uniform circular motion**describes

**motion of a body**traversing a

**circular path**at a constant speed. Since the body describes

**circular motion**, its

**distance from the axis of rotation remains constant throughout**. Although a

**body’s speed is constant**, its speed is not constant.

**V**

**elocity, a vector quantity**, depends on both the body’s

**speed**and

**direction.**

**Expert Answer**

**Given**\[OA = r \] \[OB = 2r \] \[m_{A} = m \] \[m_{B} = 2m\] \[v_{A} = 1000 \dfrac {m}{s} \] The

**speed of satellite**$ 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**$ 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**$ 2 $ is given as: \[\dfrac{g_{A}}{g_{B}} = 4. \dfrac{v_{A} ^ {2}}{v_{B} ^ {2}}\] By

**equalizing both equations**$ 1 $ and $ 2 $: \[\dfrac{m_{A}}{m_{B}} = \dfrac{v_{A} ^ {2}}{v_{B} ^ {2}}\] \[m_{A} = m\: ; \: m_{B} = 2m;\] \[\dfrac{m}{2m} = \dfrac{v_{A}^{2}}{v_{B} ^ {2}}\] \[ v_{B} = v_{A}.\sqrt {2} \] \[ v_{A} = 10000 \dfrac {m} {s} \] \[ v_{B} = 10000 \sqrt 2 \dfrac {m} {s} \]

**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

**speed of satellite**$B$ 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**$1$ 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**$2$ is given as: \[\dfrac {g_{A} }{ g_{B} } = 4. \dfrac {v_{A} ^ {2} }{v_{B} ^ {2} }\] By

**equalizing both equations**$ 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**$B$ is $20000 \sqrt 2 \dfrac {m} {s} $.

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