 # Satellite A orbits a planet with a speed of 10,000 osteoblogda 2021-12-18 Answered
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.
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Step 1
Given values:
${v}_{A}=10000\frac{m}{s}$
${m}_{B}=2{m}_{A}$
${r}_{B}={r}_{A}$
The gravitational force is equal to:
${F}_{GA}=G\frac{{m}_{A}m}{{r}_{{A}^{2}}}$ (Equation 1.)
${F}_{GB}=G\frac{{m}_{B}m}{{r}_{{B}^{2}}}$
${F}_{GB}=G\frac{2{m}_{A}m}{2{r}_{{A}^{2}}}$
${F}_{GB}=G\frac{2{m}_{A}m}{4{r}_{{A}^{2}}}$ (Equation 2.)
According to the Newtons

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${v}_{B}=10000\sqrt{2}\frac{m}{s}$
Given:
$OA=r;OB=2r$
${m}_{A}=m;{m}_{B}=2m$
${v}_{A}=10000\frac{m}{s};{v}_{B}=?$
${m}_{p}$: mass of planet
$\frac{{m}_{A}\cdot {g}_{A}}{{m}_{B}\cdot {g}_{B}}=\frac{K\cdot \frac{{m}_{A}\cdot {m}_{P}}{{r}^{2}}}{\frac{{m}_{B}\cdot {m}_{p}}{4{r}^{2}}}$
yields $\frac{{g}_{A}}{{g}_{B}}=4\cdot \frac{{m}_{A}}{{m}_{B}}\left(1\right)$
$\frac{{m}_{A}\cdot {g}_{A}}{{m}_{B}\cdot {g}_{B}}=\frac{\frac{{m}_{A}\cdot {v}_{A}^{2}}{{r}^{2}}}{\frac{{m}_{B}\cdot {v}_{B}^{2}}{4{r}^{2}}}$
yields $\frac{{g}_{A}}{{g}_{B}}=4\cdot \frac{{v}_{A}^{2}}{{v}_{B}^{2}}\left(2\right)$
so $\left(1\right)=\left(2\right)$
$\frac{{m}_{A}}{{m}_{B}}=\frac{{v}_{A}^{2}}{{v}_{B}^{2}}$
${m}_{A}=m;{m}_{B}=2m$;
$\frac{m}{2m}=\frac{{v}_{A}^{2}}{{v}_{B}^{2}};{v}_{B}^{2}=2\cdot {v}_{A}^{2};{v}_{B}={v}_{A}\cdot \sqrt{2}$
${v}_{A}=10000\frac{m}{s}$
${v}_{B}=10000\sqrt{2}\frac{m}{s}$

We have step-by-step solutions for your answer! nick1337

Satellite B's mass is not relevant, but at twice the distance, its orbital period is greater by a factor of $\sqrt{\left(2{\right)}^{3}}=2\sqrt{2}$. But the distance it must travel during each period is also greater by a factor of 2. Thus, its speed is $\frac{2}{2\sqrt{2}}$ less or $\frac{1}{\sqrt{2}}$ less.
${V}_{B}=0.7071$
${V}_{A}=7.071\frac{m}{s}\approx 7×{10}^{3}\frac{m}{s}.$

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