Question

A meteorite has a speed of 80.0 m/s when 900 km above the Earth. It is falling vertically (ignore air resistance) and strikes a bed of sand in which it is brought to rest in 3.20 m .

Answer 1

Answer:

Part a)

$v = 3950.5 m/s$

Part b)

$W = -4.56 \times 10^9 J$

Part c)

$F = 1.43 \times 10^9 N$

Part d)

$Energy = 4.56 \times 10^9 J$

Explanation:

Part a)

As we know that total mechanical energy of the meteorite must be conserved

so we will have

$-\frac{GMm}{R+ h} + \frac{1}{2}mv_i^2 = - \frac{GMm}{R} + \frac{1}{2]mv_f^2$

$-\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{6.37 \times 10^6 + 9\times 10^5} + \frac{1}{2}(80^2) = -\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{6.37 \times 10^6} + \frac{1}{2}v^2$

$-5.48 \times 10^7 + 3200 = -6.26 \times 10^7  + \frac{v^2}{2}$

$v = 3950.5 m/s$

Part b)

Work done by sand to stop it

$W = \Delta K$

$W = -\frac{1}{2}mv^2$

$W = -\frac{1}{2}(585)(3950.5^2)$

$W = -4.56 \times 10^9 J$

Part c)

As we know that

W = F. d

so we have

$-4.56 \times 10^9 = F \times (3.20)$

$F = 1.43 \times 10^9 N$

Part d)

Work done = thermal energy

$Energy = 4.56 \times 10^9 J$