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‘š/𝑠. 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‘š/𝑠 )? Give your answer to the nearest integer in π‘˜π‘š/𝑠

Solution:

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

𝐾𝐸𝑖 + 𝑃𝐸𝑖 = 𝐾𝐸𝑓 + 𝑃𝐸𝑓

(1/2) π‘šπ‘’02 + (− πΊπ‘€π‘š/10𝑅 ) = (1/2) π‘šπ‘£ 2 + (− πΊπ‘€π‘š/𝑅 )

𝑣 2 = 𝑒02 + 2𝐺𝑀 𝑅 [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‘š/𝑠𝑒𝑐2 = √ 2𝐺𝑀 𝑅

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

= √256.9 ≈ 16 π‘˜π‘š/𝑠

Answer: (16)

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