How big would my telescope have to be if I wanted to see the Mars rover from my backyard? I imagine that with a big enough telescope, I would be able to zoom in and see the Mars rover in enough detail to make out the details (like the wheels, cameras, etc.). How large would the telescope have to be? (or how can I calculate this value?)
 A: Short answer
REALLY BIG
Long answer
There are several parameters that need to be controlled in order to make a properly-functioning telescope, but arguably the two most important parameters are its magnification power and its aperture size.
Magnification is defined fairly straightforward as the ratio between how big the object looks when looking at it through the eyepiece of the telescope to how big the object looks when looking at it with the naked eye. This depends on the properties of the lenses you use in your telescope. For a fairly simple telescope, you will have two lenses: the objective lens at the front of the telescope, and the eyepiece lens, the lens you actually put your eye up to in order to look at the night sky. The magnification can be expressed in terms of each lens' focal length,
$$M = \frac{f_o}{f_e}$$
where $f_o$ is the focal length of the objective lens, and $f_e$ is the focal length of the eyepiece lens. The sum of the two focal lengths will give you a rough estimate on how long the body of the telescope needs to be. This webpage gives a bit more detail on the derivation of this equation and the intuition behind the physics of refraction in lenses.
Once you have the correct lenses for the magnification you desire, the next parameter you need to take into account is the size of the telescope's aperture -- the opening of the telescope that actually captures the light coming from the object you want to look at. If you want to see a brighter, more resolved image, you're going to want a bigger aperture. The equation determining the field of view you can resolve is given on this page in Sean E. Lake's answer. The $D$ in his equation will give you the size of the aperture you need for your telescope.
A: The above answers have excellent math, but they neglect the most important factor: your hypothetical telescope is in your backyard, on the surface of planet Earth.
No matter how big you make the mirrors or lenses, your telescope cannot see the Mars rover from Earth, because of atmospheric distortion. You know how stars "twinkle"? You know how on a hot day, sometimes the ground shimmers? The air acts like a big blurry wobbly lens, which any ground-based telescope has to look through, limiting the level of detail that is physically possible to see.
When looking 50 million km away (such as Mars at closest approach), an ideal ground-based telescope might see objects that measure many tens of kilometers across. Anything smaller would be blurred out. Adaptive optics might manage to get the resolution into single-digit kilometers, but you're trying to see things 1000x smaller than that. It's not going to happen.
p.s. As @Pere suggests in the comments, you can circumvent this problem by placing the device off-Earth, then sending the image back via radio. Congratulations, your telescope already exists!
A: Telescope resolution is all about apparent angles. From the sounds of it, the lowest resolution you'd settle for would be something capable of resolving about $1 \operatorname{cm}$ objects, right? Well, the distance between the Earth and Mars varies, depending on the time of year, from around $0.5\operatorname{AU}$ to $2.5\operatorname{AU}$ ($7.5\times 10^{10} \operatorname{m}$ to $3.7\times 10^{11} \operatorname{m}$). At those distances, a $1$ centimeter object subtends an angle of
$$\theta = \frac{s}{d},$$
which is $1.5\times 10^{-13}\operatorname{rad}$ to $2.7\times 10^{-14}\operatorname{rad}$.
The resolution of a circular telescope is given by the formula
$$\theta = \frac{1.22\lambda}{D}.$$
So, assuming you're using visible light, with $\lambda \approx 500\operatorname{nm}$, to resolve those $1$ centimeter objects it would require telescopes with a diameter of $D=4.6\times 10^6\operatorname{m}$ to $7.4\times 10^7\operatorname{m}$. For reference, the diameter of Earth is about $1.3\times 10^7\operatorname{m}$.
Note that the sheer size is only one of the challenges. In order to achieve this theoretical resolution you would need the surface of the mirror to have the correct shape everywhere to within about a wavelength of light. In other words, this Earth-sized mirror could not have any imperfections larger than about $500\operatorname{nm}$. To see some of the information related to getting ordinary lenses and mirrors correct to this level see the Wikipedia article on optically flat.
