# The Gran Canyon Diablo Crater in Arizona is 200m, and was produced by an impact of a 3xx10^8 kg meteorite traveling at 1.3xx10^4 m/s. Estimate (a) the change in Earth’s velocity as a result of the impact and (b) the average force exerted on Earth?

Jan 4, 2017

Assuming that the velocity of meteorite has been stated with respect to a reference frame in which earth is stationary, and that none of the kinetic energy of the meteorite is lost as heat sound etc., we make use of the law of conservation of momentum

(a). Noting that earth's initial velocity is $0$.
And after the collision the meteorite sticks to earth and both move with same velocity. Let final velocity of earth+meteorite combine be ${v}_{C}$. From the equation stated below we get

$\text{Initial Momentum "= " Final momentum}$
$\left(3 \times {10}^{8}\right) \times \left(1.3 \times {10}^{4}\right) = \left(3 \times {10}^{8} + 5.972 \times {10}^{24}\right) \times {v}_{C}$
where 5.972 × 10^24kg is mass of earth.

We observe that velocity of meteorite is of the order of ${10}^{4} m {s}^{-} 1$ is much smaller than velocity of earth which is of the order of ${10}^{24} m {s}^{-} 1$ hence is ignored in the denominator.

$\implies {v}_{c} \approx \frac{3 \times {10}^{8} \times 1.3 \times {10}^{4}}{5.972 \times {10}^{24}}$
$= 6.5 \times {10}^{-} 13 m {s}^{-} 1$
This is change in the speed of earth due to collision with the meteorite.
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Compare with Earth's mean orbital speed of $3.0 \times {10}^{4} m {s}^{-} 1$
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(b) We know that acceleration due to gravity $= 9.81 m {s}^{-} 2$.
Taking same as average value of acceleration acting on the meteorite,
Average force exerted on Earth $F = m g$
$\implies F = \left(3 \times {10}^{8}\right) \times 9.81 = 2.94 \times {10}^{9} N$, rounded to two decimal places.