An asteroid known as 2008 OS7, with an estimated width six times greater than the height of Delhi’s Qutub Minar, the world’s tallest minaret, is set to have a close encounter with Earth on February 2. The asteroid, believed to be between 690 and 1,570 feet wide, comparable to the Empire State Building in New York, will pass by at a distance of about 1.77 million miles. To provide context, Venus, our neighboring planet, is approximately 38 million miles away at its closest point, and the Moon orbits around 240,000 miles from Earth.
As it approaches Earth, the asteroid will be traveling at a speed of 18.2 km per second, making its closest flyby in the afternoon of February 2.
Asteroids, remnants from the formation of the solar system, are commonly found in the Asteroid Belt between Mars and Jupiter. While many are small, some, like Ceres, can be significant in size, reaching about 600 miles across. Occasionally, either due to Jupiter’s gravitational influence or collisions, these celestial bodies deviate into the inner solar system, passing by Earth.
2008 OS7 falls under the classification of both a “Near Earth Object” (NEO) and a “potentially hazardous asteroid” (PHA) because of its size and proximity to Earth. NEOs are those within 30 million miles of Earth, encompassing around 31,000 objects within the solar system. PHAs are even closer, within 4.6 million miles, and exceed 460 feet in diameter. NASA currently monitors approximately 2,350 PHAs.
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