You can use a transistor to pull down the 12V source to switch the heavy relay.
this is a circuit I used to get a 3v output to turn on a 12v relay..Make sure you use the 1n4001 diode as the coil will have alot of back emf and it hast go somewhere. I havent tested it ..but in lue of the 47K resistor start with a 100k and get rid of the 1n4148 zener diode.
The led and resistor is just there to notify me that the relay is on,also to let me know theres power to the relay if the switched current isnt working,,letting me know that the relay is shot.
for what is worth the most current Ive had drawn by a similar realay is less than 100ma..
I have a circuit that switches 16 relays in a bidirectional chaser array that use'a less than 200ma at full speed..and about 110ma at low speed.
The relay's only draw 30ma@12v..the 555 Timer can sink 25ma ..it can sink 200ma but lets not get carried away..
INCREASING OUTPUT
CURRENT
The 555 will deliver 200mA to a load but the chip gets extremely hot (12v supply). The answer is to use a buffer transistor.
For 200mA, use a BC547 or equivalent.
For 500mA use a BC337 or equivalent
For 1A, use a TIP31 or equivalent.
For 3A - 5A use a BD679 or equivalent with heatsink
For 5A to 10A use TIP3055 with heatsink This is basicly what i did in my own schematic,,using a transistor to sink the current to the load
You can fill yer boots full of 555 circuits here..
http://talkingelectronics.com/projects/50%20-%20555%20Circuits/50%20-%20555%20Circuits.html#H