Question

The gravitational force,F, on a rocket at a distance,r, from the center of the earth isgiven byF=kr2wherek= 1013N·km2. (Newton·kilometer2). When the rocket is 104km from thecenter of the earth, it is moving away at 0.4 km/sec. How fast is the gravitational forcechanging at that moment?

Answer 1

Answer:

The gravitational force changing velocity is

$\frac{dF}{dt}=-8\frac{N}{s}$

Explanation:

The expression for the gravitational force is

$F=\frac{k}{r^{2}}\\\\k=10x10^{13} N*km^{2}\\\\r=10x10^{4} km\\\\V=0.4 \frac{km}{s}$

Differentiate the above equation

$\frac{dF}{dt}=\frac{k}{r^{2}}\\\frac{dF}{dt}=k*r^{-2}\\\frac{dF}{dt}=-2*k*r^{-3} \frac{dr}{dt}\\\frac{dF}{dt}=\frac{-2k}{r^{3}}\frac{dr}{dt}$

The velocity is the distance in at time so

$V=\frac{dr}{dt}=0.4 \frac{km}{s}$

$\frac{dF}{dt}=\frac{-2*k}{r^{3}}*0.4\\\frac{dF}{dt}=\frac{-8*10x^{13}N*km^{2} }{(10x10^{4}) ^{3}} \\\frac{dF}{dt}=\frac{-8x10^{12} }{1x10^{12}}$

$\frac{dF}{dt}=-8\frac{N}{s}$