# An asteroid is moving directly towards the centre of the earth. When at a distance of 10R (R is the radius of the earth) from the earth

An asteroid is moving directly towards the centre of the earth. When at a distance of 10R (R is the radius of the earth) from the earth’s centre, it has a speed of 12 ππ/π . Neglecting the effect of earth’s atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is 12 ππ/π  )? Give your answer to the nearest integer in ππ/π

Solution:

Taking, asteroid and earth as an isolated system conserving Total energy.

πΎπΈπ + ππΈπ = πΎπΈπ + ππΈπ

(1/2) ππ’02 + (− πΊππ/10π ) = (1/2) ππ£ 2 + (− πΊππ/π )

π£ 2 = π’02 + 2πΊπ π [1 – (1/10)] $$v= \sqrt{u_{0}^{2}+\frac{9}{5}\frac{GM}{R}}$$

Since, escape velocity from surface of earth is 11.2 ππ/π ππ2 = √ 2πΊπ π

$$v= \sqrt{12^{2}+\frac{9}{5}\frac{(11.5)^{2}}{2}}$$

= √256.9 ≈ 16 ππ/π