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Very little

Earth's surface gravity is $1g$, felt at a radius of about $4,000$ miles from the planet's center. The distance to the moon is roughly $240,000$ miles, which is ~$60$ times the earth's radius. Because gravity decreases with the square of distance, the earth's pull on the moon is only $\frac{1}{3600}g$.

The moon's surface gravity is ~$\frac{1}{6}g$, and the earth's gravity will make you lighter/heavier by ~$\frac{1}{3600}g$ depending on if you're standing on the near, where the Earth is pulling away, so the gravity is $(\frac{1}{6}-\frac{1}{3600})g$ or far side of the moon, where it is $(\frac{1}{6}+\frac{1}{3600})g$. Your apparent weight will only vary by a few tenths of a percent between the near and far side of the moon.

I've ignored the radius of the moon itself here (~$1000$ miles), as it is relatively small compared to the earth-moon distance. But there is an even smaller tug from earth on the far side than the near side, further reducing the change in apparent weight.

EDIT: This answer is wrong, but will remain here as testament to incorrect reasoning. @David Hammen has it correct - the earth's pull actually decreases apparent weight on both the near and far side of the moon, much like how the moon creates a tidal bulge on both sides of the earth. The near side of the moon gets pulled more than the moon as a whole, while the moon as a whole gets pulled more than the far side.

Very little

Earth's surface gravity is $1g$, felt at a radius of about $4,000$ miles from the planet's center. The distance to the moon is roughly $240,000$ miles, which is ~$60$ times the earth's radius. Because gravity decreases with the square of distance, the earth's pull on the moon is only $\frac{1}{3600}g$.

The moon's surface gravity is ~$\frac{1}{6}g$, and the earth's gravity will make you lighter/heavier by ~$\frac{1}{3600}g$ depending on if you're standing on the near, where the Earth is pulling away, so the gravity is $(\frac{1}{6}-\frac{1}{3600})g$ or far side of the moon, where it is $(\frac{1}{6}+\frac{1}{3600})g$. Your apparent weight will only vary by a few tenths of a percent between the near and far side of the moon.

I've ignored the radius of the moon itself here (~$1000$ miles), as it is relatively small compared to the earth-moon distance. But there is an even smaller tug from earth on the far side than the near side, further reducing the change in apparent weight.

EDIT: This answer is wrong, but will remain here as testament to incorrect reasoning. @David Hammen has it correct - the earth's pull actually decreases apparent weight on both the near and far side of the moon, much like how the moon creates a tidal bulge on both sides of the earth. The near side of the moon gets pulled more than the moon as a whole, while the moon as a whole gets pulled more than the far side. The overall scale of the effects is correct though, with the earth only exerting a few ten-thousandths of a g at that distance.

Very little

Earth's surface gravity is $1g$, felt at a radius of about $4,000$ miles from the planet's center. The distance to the moon is roughly $240,000$ miles, which is ~$60$ times the earth's radius. Because gravity decreases with the square of distance, the earth's pull on the moon is only $\frac{1}{3600}g$.

The moon's surface gravity is ~$\frac{1}{6}g$, and the earth's gravity will make you lighter/heavier by ~$\frac{1}{3600}g$ depending on if you're standing on the near, where the Earth is pulling away, so the gravity is $(\frac{1}{6}-\frac{1}{3600})g$ or far side of the moon, where it is $(\frac{1}{6}+\frac{1}{3600})g$. Your apparent weight will only vary by a few tenths of a percent between the near and far side of the moon.

I've ignored the radius of the moon itself here (~$1000$ miles), as it is relatively small compared to the earth-moon distance. But there is an even smaller tug from earth on the far side than the near side, further reducing the change in apparent weight.

Very little

Earth's surface gravity is $1g$, felt at a radius of about $4,000$ miles from the planet's center. The distance to the moon is roughly $240,000$ miles, which is ~$60$ times the earth's radius. Because gravity decreases with the square of distance, the earth's pull on the moon is only $\frac{1}{3600}g$.

The moon's surface gravity is ~$\frac{1}{6}g$, and the earth's gravity will make you lighter/heavier by ~$\frac{1}{3600}g$ depending on if you're standing on the near, where the Earth is pulling away, so the gravity is $(\frac{1}{6}-\frac{1}{3600})g$ or far side of the moon, where it is $(\frac{1}{6}+\frac{1}{3600})g$. Your apparent weight will only vary by a few tenths of a percent between the near and far side of the moon.

I've ignored the radius of the moon itself here (~$1000$ miles), as it is relatively small compared to the earth-moon distance. But there is an even smaller tug from earth on the far side than the near side, further reducing the change in apparent weight.

EDIT: This answer is wrong, but will remain here as testament to incorrect reasoning. @David Hammen has it correct - the earth's pull actually decreases apparent weight on both the near and far side of the moon, much like how the moon creates a tidal bulge on both sides of the earth. The near side of the moon gets pulled more than the moon as a whole, while the moon as a whole gets pulled more than the far side.

Very little

Earth's surface gravity is 1g, felt at a radius of about 4,000 miles from the planet's center. The distance to the moon is roughly 240,000 miles, which is ~60 times the earth's radius. Because gravity decreases with the square of distance, the earth's pull on the moon is only 1/3600g.

The moon's surface gravity is ~1/6g, and the earth's gravity will make you lighter/heavier by ~1/3600g depending on if you're standing on the near or far side of the moon. Your apparent weight will only vary by a few tenths of a percent between the near and far side of the moon.

I've ignored the radius of the moon itself here (~1000 miles), as it is relatively small compared to the earth-moon distance. But there is an even smaller tug from earth on the far side than the near side, further reducing the change in apparent weight.

Very little

Earth's surface gravity is $1g$, felt at a radius of about $4,000$ miles from the planet's center. The distance to the moon is roughly $240,000$ miles, which is ~$60$ times the earth's radius. Because gravity decreases with the square of distance, the earth's pull on the moon is only $\frac{1}{3600}g$.

The moon's surface gravity is ~$\frac{1}{6}g$, and the earth's gravity will make you lighter/heavier by ~$\frac{1}{3600}g$ depending on if you're standing on the near, where the Earth is pulling away, so the gravity is $(\frac{1}{6}-\frac{1}{3600})g$ or far side of the moon, where it is $(\frac{1}{6}+\frac{1}{3600})g$. Your apparent weight will only vary by a few tenths of a percent between the near and far side of the moon.

I've ignored the radius of the moon itself here (~$1000$ miles), as it is relatively small compared to the earth-moon distance. But there is an even smaller tug from earth on the far side than the near side, further reducing the change in apparent weight.