Binary planet
http://farm1.staticflickr.com/164/358228043_d07837333f_z.jpg?zz=1![binary planet](https://i.sstatic.net/JnPXq.jpg)
Our own moon has enough gravity to tug up a bulge in the ocean, producing tides. But what if our planet's partner were even bigger? If the partner were as big or even bigger than the planet in question, its gravity could be felt by people on the ground. When overhead it would pull on objects to some degree, effectively making them lighter. When on the far side its pull would be additive to that of the planet, making things heavier.
One would need to work jiggery pokery with the masses, distances and velocities of these 2 planets such that the orbital period was as requested by the OP and the mass of the companion planet was correct to make the 20%-30% decrease when overhead. The OP did not ask for a 20-30% increase in gravity also but that is what she gets.
As regards noticeable, a giant planet sized satellite would be noticeable to people on the planet's surface. I am sure basketball season would be timed for when it is high in the sky.
ADDENDUM
from comments
I'm trying to crunch the numbers for this scenario, and I'm failing to
get any "sane" results - may be an error with my calculations, though
- @Will (or anyone else, for that matter), would you mind suggesting distance and mass for the satellite that produces the 20-30% reduction
in perceived surface gravity (let's ignore the seasonality of the
effect for simplicity)? – G0BLiN 2 days ago
OK @G0BLiN with your tricky to spell username. Math is not my strong suit. But there are calculators to help. I used this one.
https://www.omnicalculator.com/physics/gravitational-force
First: Earth Plugging in my own 100 kg weight standing at Earths radius 6371 (6371 km from the center of mass) I get:
981.7 N / 100 kg = 9.8 m/s2 which is Earths surface gravity. Looks good.
Now let us consider Jupiter at 317 Earth masses and radius 69911 km. How far off the surface can on object in orbit exist? Metis is the closest moon of Jupiter, orbiting at 128,000 km. 128000 + 69911 = 197911 km from Jupiter’s center of mass.
![enter image description here](https://i.sstatic.net/0nYIF.jpg)
322.5 N / 100 kg = 3.2 m/s2. Which is about a third of Earth’s gravity.
So: if I were standing on Earth’s surface looking at Jupiter 197911 km overhead, I would be lifted up by Jupiter with a force of 30% of Earth’s own pull. I would feel 30% lighter. On the far side of Earth I have the diameter of the Earth between us too; if it is 12742 then the force is 2.8 N in addition to the 9.8 of earth.
A problem: to orbit as closely as Metis does a body must move very fast. It is falling towards Jupiter at a good clip and it has to reliably miss. A fix: the Earth in this scenario orbits in a highly elliptical orbit.
Ha! Gifs work on WB stack!
This can address the periodicity also – for much of the 18 month orbit the Earthlike body is at some distance from its large Jupiteresque partner. It swoops close for the specified period in the OP during which the proximity allows the above described gravity reduction due to the large partner.
[distance from primary] < 0.2x[min distance from secondary]
(see here)), but for that minimal distance, the secondary must be around x10,000 more massive than the sun (the primary) to negate 30% of the planet's gravitational pull. This makes the secondary a super-massive black hole, which (probably?) makes the whole system non-viable... $\endgroup$