The escape velocity is the minimum speed an object needs to reach to escape the gravitational pull of a celestial body. On Earth, the escape velocity is approximately 11.2 kilometers per second (6.95 miles per second) at the surface. However, as an object moves farther away from Earth, the gravitational pull weakens, and the escape velocity decreases.
To calculate the radius at which the escape velocity becomes equal to the speed of light, we can use the formula for escape velocity:
v = √(2GM/r)
Where: v is the escape velocity G is the gravitational constant (approximately 6.67430 × 10^(-11) m^3 kg^(-1) s^(-2)) M is the mass of Earth (approximately 5.972 × 10^24 kg) r is the distance from the center of Earth
We want to find the radius (r) at which the escape velocity (v) equals the speed of light (c), which is approximately 299,792 kilometers per second (299,792,458 meters per second).
Setting v equal to c, we have:
c = √(2GM/r)
Squaring both sides of the equation, we get:
c^2 = 2GM/r
Rearranging the equation to solve for r, we have:
r = 2GM/c^2
Now we can substitute the known values and calculate the radius:
r = (2 * 6.67430 × 10^(-11) m^3 kg^(-1) s^(-2) * 5.972 × 10^24 kg) / (299,792,458 m/s)^2
Calculating this expression gives us:
r ≈ 8.8708 × 10^26 meters
Converting this to kilometers, we have:
r ≈ 8.8708 × 10^23 kilometers
Therefore, the radius at which the escape velocity from Earth becomes equal to the speed of light is approximately 8.8708 × 10^23 kilometers.