Question

On Earth, froghoppers can jump upward with a takeoff speed of 2.8 m/s. Suppose you took some of the insects to an asteroid. If it is small enough they can jump free of it and escape into space. (a) What is the diameter (in km) of the largest spherical asteroid from which they could jump free? Assume typical asteroid denisty of 2.0 g/cm^3. (b) Suppose that one of the froghoppers jumped horizontally from a small hill on an asteroid. What would the diameter (in km) of the ateroid need to be so that the insect could go into a circular orbit just above the surace?

Answer 1

To develop this problem it is necessary to apply the concepts related to kinetic energy and gravitational potential energy.

By conserving energy we know that

$PE = KE$

$\frac{1}{2}mv^2=\frac{GMm}{r}$

Where,

m = mass

v = Velocity

G = Gravitational universal constant

M = Mass of Spherical asteroid

m = mass of object

r = Radius

Mass of a Sphere can be expressed as,

$M= \rho* (\frac{4}{3}\pi r^3 )$

Replacing we have that,

$\frac{1}{2}*2.8^2 - 6.67*10^{-11}*\frac{4\pi*2000*r^3}{3r} = 0$

$6.67*10^{-11}*(\frac{4\pi}3{}2000*r^2)=\frac{1}{2}*2.8^2$

$r = \sqrt{\frac{1}{2}*\frac{2.8^2}{6.67*10^{-11}*2000*4/3*\pi}}$

$r=2648 m$

Therefore the diameter is 5296 m.

b) Applying the concept of gravitational force and centripetal force we have to

$F_g = F_c$

$\frac{G M m}{r^2} = \frac{m v^2}{r}$

$\frac{G M}{r} = v^2$

$6.67*10^{-11}*\frac{(\frac{4\pi}{3}2000*r^3)}{r} = 2.8^2$

$r=3746 m$

Therefore the diameter would be 7492 m