Satellite Speed

Relevant wiki: Gravitation

An object moving in circular motion will need a force towards center i.e centripetal force. Here it is being provided by gravity. $\dfrac{mv^2}{r} = GM\dfrac{m}{r^2}$
On solving this we get $v \propto \dfrac{1}{\sqrt{r}}$
This means $\color{#D61F06}{\text{To maintain a constant height at a lower orbit, a higher speed is needed}}$
Hence answer is 'It will increase'.

We know that the velocity of a spacecraft at a radius $r$ from the earth and the mass of earth being $M_E$ is given by the expression:

$v=\sqrt{G\dfrac{M_E}{r}}$

Now, let us say that the spacecraft was initially orbiting at a radius $r_1$ with a velocity of $v_1$ and it moves to a lower orbit of radius $r_2$ and changes the velocity to $v_2$ . We will compute the change in velocity as the spacecraft moves from $r_1$ to $r_2$ :

$\begin{aligned} \Delta v &= v_2-v_1\\ &= \sqrt{G\dfrac{M_E}{r_2}}-\sqrt{G\dfrac{M_E}{r_1}}\\ &=\sqrt{GM_E\left(\dfrac{1}{r_2}-\dfrac{1}{r_1}\right)}\\ &=\sqrt{GM_E\left(\dfrac{r_1-r_2}{r_1r_2}\right)}\\ \end{aligned}$

Thus we can observe that there is a positive change in the velocity of the spacecraft, which means that the velocity increases.

Consider the total mechanical energy in the process,
as there is no external forces,
Kinetic Energy + Gravtional Potential Energy is a constant.
Moving to a lower orbit is a drop in GPE, so KE increases.
KE increases means speed increases.

Kepler's Third Law states that the two orbits compare as $\frac{r^3}{T^2} = \text{const}.$ With $v \propto r/T$ , this becomes $r\cdot v^2 = \text{const},$ so that $v \propto \frac 1{\sqrt r}.$ Smaller orbits will require higher speeds.