Because you want to control the amplification of a signal that is DC (maybe below 100Hz is my guesstimate), "conventional" devices such as JFETs are probably going to be fairly poor in DC performance. This would also apply to analogue multipliers such as the AD534 because temperature changes do affect analogue multipliers.
A better solution is to control an analogue switch from a square wave whose mark-space ratio represents the analogue control voltage. Your input feeds the switch and the output from the switch is your output.
If the square wave (say 10kHz) alternates between 0V (50% of the time) and 5V (50% of the time) and this switched the input signal (Vin) "on" and "off" via an analogue switch, the average output would be 50% of Vin. If the on/off ratio were altered to say 20% on (5V) and 80% off (0V), the average output would be 20% of Vin. Why do it this way?
It's quite easy to get a very accurate and reliable on/off duty-cycle using either analogue electronics or digital electronics and using this to "chop" the input to control its amplitude is a guaranteed way of overcoming the JFET/Multiplier problems associated with DC and temperature drift.
The output of the analogue switch would of course be "chopped-up" but the average value could entirely represent the transfer function you need. In addition, you can use standard filter techniques to recover the chopped signal back to a steady dc version.
You probably noticed that I mentioned the input being limited to 100Hz, and that I mentioned a chopping frequency of 10kHz - I did this purposely because this allows for a simple filter arrangement to recover the chopped output back to a smooth dc level.
Here's a simulation of a switching circuit: -
The output waveform (BLUE) is half a cycle of an input sinewave (100Hz) chopped by 10kHz 50:50 duty-cycle The chopping is done by the analogue switches S1 and S2. These are, of course, common-place devices. The red trace is the blue trace having undergone filtering by the op-amp circuit.
Because the duty-cycle of the square wave controlling the switches is 50:50, the output signal is halved in value. If the duty cycle were 1:99 (on for 1% of the time) then the output would be one-hundredth of the input.
This is how I'd tackle the problem BUT you could always use a cheap MCU and feed both inputs into its ADC and get the MCU to calculate the output value and have this fed to a DAC.
You want to measure dc voltages up to say 15 volts and these might have 2 volts of ripple superimposed for a half guess. Your smallest dc voltage might be 3v and you think your measurement accuracy may be unduly affected at the low end, yes?
Don't worry about it - make your input suitable for top peaks of 20 volts and you'll find that with a ten bit ADC you'll be just fine. Ten bits means a resolution of 20 ish milli volts and with a smidgen of noise superimposed and several samples averaged you'll easily achieve decent accuracy. Taking two samples then averaging gives you an 11 bit number and this will be accurate to 11 bits with out of band ac noise on top.
Try looking up dithering - it's a well formulated technique that improves the resolution of 16 bit CD audio so that it sounds less grainy to the ear on quiet passages of music. You just need a little bit of out of band noise to make it work.
Best Answer
You'll need 5 dB overall gain drop, so your options here seem to be the following ones:
SIMPLEST (and straightforward) OPTION): use an external 50 \$\Omega\$ 5 dB attenuator capable of handling 8W input power. Design and build it from scratch using one of the many available online calculators.
SIMPLE (but not straighforward) OPTION: Increase the input attenuation by 5 dB. This means changing the values of \$R_2\$, \$R_3\$, \$R_4\$ and \$R_5\$, and also means changing the power rating of the resistors because they will dissipate more power. The input attenuator also provides impedance matching, so you should recalculate these values taking into account \$R_8\$ and the input impedance of the MOSFET gate (according to the datasheet: 180 pF @ 1 MHz).
COMPLEX OPTION: Reduce the gain of the transistor by 5 dB (this is is also a more power-efficient option). You achieve this by changing the P1 potentiometer setting so that the MOSFET gate voltage is lower and the bias current is reduced. You can adjust the gain to a lower setting during the tuning process. In this case you don't have to change the values of \$R_2\$, \$R_3\$, \$R_4\$ and \$R_5\$, but you must change their power rating along with the power rating of \$R_8\$ and the current rating of \$L_1\$. You must also check that the gate-source voltage doesn't exceed the +/20V voltage rating of MOSFET. At 8W input, this a very real possibility and could mean this option is not feasible.
EVEN MORE COMPLEX (but possibly optimal) OPTION: A combination of option 2 and 3. Split the 5 dB gain drop between the input attenuator and the transistor gain itself.
For reference purposes: IRF510 datasheet.