91Ó°ÊÓ

Given that the radius \(r\) of the object is \(50\;{\rm{m}}\), the net force \({F_{{\rm{net }}}}\) achieved by the rocket is \(3.4 \times {10^7}\;{\rm{N}}\), the time interval \(\Delta t\) which rocket burns is \(170\;{\rm{s}}\), assume that the initial velocity \({v_i}\) of the object is \(3 \times {10^5}\;{\rm{m}}/{\rm{s}}\) and assume that the density \(\rho \) of the rock is \(8 \times {10^3}\;{\rm{kg}}/{{\rm{m}}^3}\).

The quantity of motion that occurs in something that is moving, or the force that propels something forward to keep it moving, is defined as momentum.

The rate at which an automobile descends a hill is an illustration of momentum.

Formulae of momentum is\(p = mv\).

calculate the mass of this object,

\(\)\(\begin{aligned}{c}m &= \rho V\\ \approx \left( {7.5 \times {{10}^3}} \right)\left( {\frac{4}{3} \times \pi \times {{50}^3}} \right)\\ &= 3.93 \times {10^9}\;{\rm{kg}}\end{aligned}\)

Utilize the momentum principle when the rocket and the object collide, and the impulse will \(\)be

\(\begin{aligned}{c}{\rm{ Impulse }} &= F\Delta t\\ &= \left( {3.4 \times {{10}^7}} \right)(170)\\ &= 5.78 \times {10^9}\;\;{\rm{N}} \cdot {\rm{s}}\end{aligned}\)

Thus, the object will have a speed of

\(\begin{aligned}{c}{v_f} &= \frac{{{p_f}}}{m}\\ &= \frac{{1.18 \times {{10}^{14}}}}{{3.93 \times {{10}^9}}}\\ \approx 3 \times {10^4}\;{\rm{m/s}}\end{aligned}\)

The object will have a speed of \(3 \times {10^4}\;{\rm{m}}/{\rm{s}}\)

Thus, the objects' initial and ultimate speeds become comparable. Because the most powerful rocket ever fired has no effect on the object's speed, it will arrive at the point where the object collides with the ground at the same time. According to the assumptions made in the answer, the earth would not be preserved. As a result, it is uncertain.