Schwarzschild Radius
Radius of the event horizon of a black hole.
Formula
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Example
Sun would be ~3 km
How it works
This formula gives the radius at which an object of mass M would have an escape speed equal to the speed of light, so light itself could no longer get out. It comes from combining Newtonian gravity with the idea that the gravitational binding at that radius is strong enough that c becomes the limiting speed, which is why the result marks a black hole’s event horizon in the simplest non-rotating case. For any spherically symmetric mass distribution compressed inside this radius, the exterior spacetime behaves as though the mass were concentrated at the center, leading to the Schwarzschild solution in general relativity.
When to use it
Use this when estimating whether a mass compressed to a given size would form a non-rotating black hole, or when converting a black hole’s mass into the size of its event horizon. It is also useful in astrophysics for quick order-of-magnitude comparisons between different black holes.
Worked example, step by step
Find the Schwarzschild radius of Earth. Given: G = 6.67430×10^-11 m^3 kg^-1 s^-2, M = 5.972×10^24 kg, c = 2.99792458×10^8 m/s Step 1: Write the formula: r_s = 2GM / c^2 Step 2: Substitute values: r_s = 2(6.67430×10^-11)(5.972×10^24) / (2.99792458×10^8)^2 Step 3: Calculate the numerator: 2GM ≈ 7.97×10^14 Step 4: Calculate the denominator: c^2 ≈ 8.99×10^16 Step 5: Divide: r_s ≈ (7.97×10^14) / (8.99×10^16) ≈ 8.87×10^-3 m Final answer: r_s ≈ 0.0089 m, or about 8.9 mm
Common mistakes to avoid
- Using the formula for any black hole without noting it applies to the non-rotating, uncharged Schwarzschild case
- Forgetting to square c, which makes the radius far too large
- Mixing units for mass or constants, such as using km or solar masses without converting consistently
Did You Know?
The Schwarzschild radius of the whole Earth is under a centimetre — that is how compressed a black hole must be.
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