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

The gravitational force changing velocity is

`(dF)/(dt)=-8(N)/(s)`

Step-by-step explanation:

The expression for the gravitational force is

`F=(k)/(r^(2))\\\\k=10x10^(13) N*km^(2)\\\\r=10x10^(4) km\\\\V=0.4 (km)/(s)`

Differentiate the above equation

`(dF)/(dt)=(k)/(r^(2))\\(dF)/(dt)=k*r^(-2)\\(dF)/(dt)=-2*k*r^(-3) (dr)/(dt)\\(dF)/(dt)=(-2k)/(r^(3))(dr)/(dt)`

The velocity is the distance in at time so

`V=(dr)/(dt)=0.4 (km)/(s)`

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

`(dF)/(dt)=-8(N)/(s)`