A projectile is shot straight up from the surface of Earth by means of a very powerful cannon and experiences no appreciable air resistance.If the projectile reaches a height of 35,000 m above Earth's surface,what was the speed of the projectile when it left the cannon? (G = 6.67 × 10^-11 N ∙ m²/kg²,the radius of Earth is 6.38 × 10⁶ m,and M_earth = 5.97 ∙ 10²⁴ kg) A)355 m/s B)505 m/s C)827 m/s D)710 m/s E)906 m/s

Question

Quizplus · Accepted Answer

To find the initial speed of the projectile, we can use the conservation of mechanical energy principle. The initial kinetic energy plus the initial potential energy equals the final potential energy (since at the highest point, the kinetic energy is 0). The formula for kinetic energy is $ \frac{1}{2}mv^2 $ and for gravitational potential energy is $ -\frac{G M m}{r} $, where $ G $ is the gravitational constant, $ M $ is the mass of the Earth, $ m $ is the mass of the projectile, $ r $ is the distance from the center of the Earth, and $ v $ is the velocity of the projectile.Initial energy (kinetic + potential) at the surface of the Earth:$$ \frac{1}{2}mv^2 - \frac{G M m}{R} $$Final potential energy at the highest point:$$ -\frac{G M m}{R + h} $$Setting the initial energy equal to the final energy and solving for $ v $ gives:$$ \frac{1}{2}mv^2 - \frac{G M m}{R} = -\frac{G M m}{R + h} $$$$ \frac{1}{2}v^2 = \frac{G M}{R} - \frac{G M}{R + h} $$$$ v^2 = 2G M \left( \frac{1}{R} - \frac{1}{R + h} \right) $$Plugging in the given values ($ G = 6.67 \times 10^{-11} $ N·m²/kg², $ M = 5.97 \times 10^{24} $ kg, $ R = 6.38 \times 10^6 $ m, $ h = 35000 $ m), we get:$$ v^2 = 2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24} \left( \frac{1}{6.38 \times 10^6} - \frac{1}{6.38 \times 10^6 + 35000} \right) $$$$ v \approx 827 \, \text{m/s} $$Therefore, the correct answer is C) 827 m/s.

A Projectile Is Shot Straight Up from the Surface of Earth