A spacecraft leaves the Earth's surface at 13 km/s…

Contributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A spacecraft leaves the Earth's surface at $13\ \text{km/s}$ . Ignoring air and the Sun, how fast is it moving very far from the Earth? What launch speed would leave it with $5.0\ \text{km/s}$ far away? (Escape speed $11.2\ \text{km/s}$ .)

Need a reminder? The key ideaShow

Escape speed: $\tfrac12 mv_e^2 = \dfrac{GMm}{R}$ , so $v_e = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR}$ , $11.2\ \text{km/s}$ for the Earth. Launched at $v > v_e$ , a body keeps $v_\infty = \sqrt{v^2 - v_e^2}$ far away; at $v < v_e$ straight up, it rises to $h = \dfrac{Rv^2}{2gR - v^2}$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

← All Problems

← Gas molecules move with a typical speed vrms = √(3kT/m), with k = 1.38 × 10⁻²³ J/K…A satellite orbits 350 km above the Earth… →

More onMechanics › Gravitation Astrophysics

Discussion (0)

Loading…