# Satellite A orbits a planet with a speed of 10,000 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.Satellite A orbits a planet with a speed of 10,000 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.

This 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. Velocity, a vector quantity, depends on both the body’s speed and direction.

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 $20000 \sqrt 2 \dfrac {m} {s}$.