Satellites In Orbit

Let $r$ be the distance from B to the center C of the Earth. Then, $2r$ is the distance from A to C. Thus, the acceleration of rocket B is $\dfrac{v_B^2}{r},$ and the acceleration of rocket A is $\dfrac{v_A^2}{2r}.$ Therefore $\begin{aligned} \dfrac{v_B^2}{r} &= \dfrac{v_A^2}{2r} \\ v_B^2 &= \dfrac{v_A^2}{2} \\ \dfrac{v_B^2}{v_A^2} &= \dfrac{1}{2} \end{aligned}$ so $\dfrac{v_B}{v_A} = \sqrt{\dfrac{1}{2}} \approx \boxed{0.707}.$

The centripedal acceleration is $\dfrac{v^2}{r}$ Let the radii of the orbits of the satellites be $r_A$ and $r_B$ . Then, $r_A=2r_B\implies\dfrac{r_B}{r_A}=\dfrac{1}{2}$ $\dfrac{{v_A}^2}{r_A}=\dfrac{{v_B}^2}{r_B}\implies\dfrac{{v_B}^2}{{v_A}^2}=\dfrac{r_B}{r_A}\implies \dfrac{v_B}{v_A}=\sqrt{\dfrac{r_B}{r_A}}=\sqrt{\dfrac{1}{2}}\approx\boxed{0.707}$

We can use the formula $a_c = \frac{v^2}{r}$ where $a_c$ , $v$ and $r$ are the centripetal acceleration, velocity and radius respectively. Let $v_A$ and $v_B$ be the velocities of sattellite A and B respectively. Since they both have the same centripetal acceleration, it can be written that $\frac{v_A^2}{2r} = \frac{v_B^2}{r}$ $\frac{v_A^2}{2} = v_B^2$ $\frac{1}{2} =\frac{v_B^2}{v_A^2}$ $\frac{v_B}{v_A} = \frac{1}{\sqrt{2}} \approx \fbox{0.707}$

Using the formula for the orbital velocity of any satellite: $v= \sqrt{gR}$ , were g is gravitational acceleration and R is distance from Earth's centre.
Given that
$R_{A} = 2R_{B}$
Now, for the first rocket:
$v_{A} = \sqrt{gR_{A}}$
or $v_{A} = \sqrt{2gR_{B}}$ .................(1)
and for the second rocket:
$v_{B} = \sqrt{gR_{B}}$ .....................(2)
So, by dividing eq2 by eq1,
$\frac{v_{B}}{v_{A}} = \sqrt\frac{gR_{B}}{2gR_{B}}$
$\frac{v_{B}}{v_{A}} = \frac{1}{\sqrt{2}}$
Therefore, $\frac{v_{B}}{v_{A}} = \boxed{0.707}$

$a_{B} = a_{A}$

$v_{B}^{2} / R_{B} = v_{A}^{2} / R_{A}$

we'll have :

$v_{B}/v_{A} = (R_{B} / R_{A})^{1/2}$ which $R_{A} = 2 R_{B}$

finally we'll have :

$v_{B}/v_{A} = 1/ √2 = 0.707$

$a_{B}=a_{A}$ $\frac{(v_{B})^{2}}{r_B}=\frac{(v_{A})^{2}}{r_A}$ $\frac{(v_{B})^{2}}{(v_{A})^{2}}=\frac{r_B}{r_A}$ in the problem it says that rocket A is twice as far from the center of the Earth as rocket B, so our equation will be $\frac{(v_{B})^{2}}{(v_{A})^{2}}=\frac{r_B}{2(r_B)}$ $\frac{v_{B}}{v_{A}}=\sqrt{\frac{1}{2}}$ $\frac{v_{B}}{v_{A}}=0.707...$