Washington, Oct 7 (IANS): An asteroid the size of a Boeing 747 airplane is set to cross the Earth's orbit at around 10.42 p.m. Indian Standard Time (IST) on Wednesday, according to NASA which expects that the space rock will not cause any harm to our home planet.

As it crosses Earth's orbit, the near-Earth asteroid named 2020 RK2 will be about 38,30,238 kilometres from Earth.

In terms of size, the asteroid is about 80 metres wide (which can be as big as Boeing 747) and travelling at 6.68km every second.

Asteroid 2020 RK2 will be hurtling through space at a speed of 6.68 kilometres per second, which is the equivalent to 14,942 miles per hour.

NASA estimates 2020 RK2 to be anywhere between 36m and 81m in diameter, and the equivalent to 118 to 265 foot wide.

It is highly unlikely keen astronomers will be able to see the asteroid from Earth, but the space rock will zip past at 1:12 pm Eastern Standard Time or 6:12 pm British Summer Time, according to dailystar.co.uk.

"NASA is tasked with discovering and tracking near-Earth #asteroids. Deep space survey and tracking technology is an area of collaboration between @NASA and @SpaceForceDOD," NASA Asteroid Watch, the Planetary Defence Coordination Office of the US space agency, said in a tweet.

NASA last month announced a new memorandum of understanding with the US Department of Defence that commits the two organisations to broad collaboration in areas including human spaceflight, US space policy, space transportation, standards and best practices for safe operations in space, scientific research, and planetary defence.