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If orbital velocity is reduced when we want to put a satellite in a higher orbit, and if to achieve a lower orbit we need to increase it's velocity, then how come by increasing the speed of the satellite we can escape the same satellite from earths gravity?

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This is an extension of Dale's answer. We need to introduce a bit of orbital mechanics. The specific energy of the satellite has the form $$ E = \frac{v^2}{2} - \frac{k}{r}.\tag{1} $$ In general, a bound orbit has the shape of an ellipse with semi-major axis $a$ and eccentricity $\varepsilon$. The distance of the satellite at its periapsis is $r_\text{peri} = a(1-\varepsilon)$, and likewise at its apoapsis $r_\text{apo} = a(1+\varepsilon)$. At the apsides, the specific angular momentum $h$ is simply the product of the distance and the velocity: $$h = r_\text{peri}v_\text{peri} = r_\text{apo}v_\text{apo}.\tag{2}$$ If we plug this into $(1)$, we get $$ E = \frac{h^2}{2a^2(1-\varepsilon)^2} - \frac{k}{a(1-\varepsilon)} = \frac{h^2}{2a^2(1+\varepsilon)^2} - \frac{k}{a(1+\varepsilon)}.\tag{3} $$ From this, we obtain $$ E\left[a^2(1+\varepsilon)^2-a^2(1-\varepsilon)^2\right] = -k\left[a(1+\varepsilon) - a(1-\varepsilon)\right],\tag{4} $$ which can be simplified to $$ E = -\frac{k}{2a}.\tag{5} $$ Plug this into $(1)$, and we have an expression of the velocity in terms of the distance and semi-major axis: $$ v^2 = \frac{2k}{r} - \frac{k}{a}.\tag{6} $$ Now, suppose we start with a satellite on a circular orbit with radius $r_1$ and velocity $v_1$. Then we have $a_1 \equiv r_1$ and $$ v_1^2 = \frac{2k}{r_1} - \frac{k}{a_1} = \frac{k}{r_1}.\tag{7} $$ We would like to bring this into a higher circular orbit with radius $r_2 > r_1$ and velocity $v_2$. For such an orbit $a_2 \equiv r_2$ and $$ v_2^2 = \frac{2k}{r_2} - \frac{k}{a_2} = \frac{k}{r_2}.\tag{8} $$ Clearly, $v_2 < v_1$. But how can we put the satellite into such an orbit? The answer: by giving it two boosts, one at distance $r_1$, and one at distance $r_2$.

First we boost it in such a way that the orbit changes from a circular orbit into an elliptical orbit with periapsis $r_1$ and apoapsis $r_2$. In other words, the new semi-major axis $\bar{a}$ and eccentricity $\bar{\varepsilon}$ must be such that $$ \begin{align} \bar{a}(1-\bar{\varepsilon}) &= a_1 = r_1,\\ \bar{a}(1+\bar{\varepsilon}) &= a_2 = r_2.\tag{9} \end{align} $$ we find $$ \begin{align} 2\bar{a} &= r_1 + r_2,\\ \bar{\varepsilon} &= \frac{r_2-r_1}{r_1 + r_2}.\tag{10} \end{align} $$ The satellite will follow this new orbit if we boost its initial velocity $v_1$ to a new velocity $\bar{v}_1$, given by $$ \bar{v}_1^2 = \bar{v}_\text{peri}^2 = \frac{2k}{r_1} - \frac{k}{\bar{a}} = \frac{r_2}{r_1}\frac{2k}{r_1 + r_2} = v_1^2\frac{2r_2}{r_1 + r_2}.\tag{11} $$ When the satellite has completed half an orbit, it will be at its apoapsis $r_2$ with velocity $\bar{v}_2$, given by $$ \bar{v}_2^2 = \bar{v}_\text{apo}^2 = \frac{2k}{r_2} - \frac{k}{\bar{a}} = \frac{r_1}{r_2}\frac{2k}{r_1 + r_2} = v_2^2\frac{2r_1}{r_1 + r_2}.\tag{12} $$ Finally, at $r_2$ we perform a second boost from velocity $\bar{v}_2$ to $v_2$, and the orbit of the satellite will change into a circular orbit with radius $r_2$. As you can see, $$\bar{v}_2 < v_2 < v_1 < \bar{v}_1,\tag{13}$$ so $\Delta v_1 = \bar{v}_1 - v_1 > 0$ and $\Delta \bar{v}_2 = v_2 - \bar{v}_2 > 0$, but $v_2 < v_1$.

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    $\begingroup$ Perfect bro..Thank you so much.. $\endgroup$ – Prashant Oct 9 at 5:41
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If orbital velocity is reduced when we want satellite to put in upper or higher orbit and if to achieve lower orbits we need to increase it's velocity, then how come is that by increasing speed of satellite we can escape the same satellite from earths gravity..??

The key difference is the shape of the orbit. For circular orbits the higher the orbit the slower the velocity, but an escape orbit is hyperbolic, not circular.

Even though a higher circular orbit is slower if you want to transition from one circular orbit to a higher circular orbit you still need to increase your speed. Increasing your speed puts you into an elliptical orbit. Then, when you reach the new orbit you need to increase your speed again to get into a circular orbit at the new altitude. Despite increasing your speed twice, you will be going slower at this new orbit.

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  • $\begingroup$ Its quite interesting though couldn't understand fully..But got an idea..Thanx..I will also be happy if you suggest some useful link explaining this point further . $\endgroup$ – Prashant Oct 8 at 11:15

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