A measure of curvature itself is:
K = 1/R^2
= GM/(c^2 r^3)
≈ 1.707 × 10^{-23}
The sum of the angles of a triangle with area A in a space with curvature K is about π + K A, so if space at a fixed moment in time has a similar curvature to the spacetime curvature we’ve found, then an equilateral triangle with sides as big as the radius of the Earth would have an angular excess or deficit of about:
K A ≈ (1.707 × 10^{-23}) × (1.76 × 10^{13})
≈ 3 × 10^{-10} radians
That would not be easy to measure directly! But we can still quantify the curvature of the spacetime around the Earth using nothing but T, the period of a surface-grazing satellite, and c, the speed of light.
Greg Egan
npub1qq…j5nhv
2026-01-11 10:53:03 UTC
in reply to nevent1q…0eg2
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npub1qqm7nu2qf25xd3mw6y6cyp4v2wr7ktf43ymp5wqz4u8jvz76ymtsjj5nhvPublished at
2026-01-11 10:53:03 UTCEvent JSON
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