A small space probe of mass 170 kg is launched from a spacecraft near Mars. It travels toward the surface of Mars, where it will eventually land. At a time 22.9 seconds after it is launched, the probe is at location <5600, 7200, 0> m, and at this same instant its momentum is <51000, -7000, 0> kg·m/s. At this instant, the net force on the probe due to the gravitational pull of Mars plus the air resistance acting on the probe is <-4000, -780, 0> N.

Assuming that the net force on the probe is approximately constant during this time interval, what is the change of the momentum of the probe in the time interval from 22.6 seconds after the probe is launched to 22.9 seconds after the launch?

The change  in momentum is        

Explanation:

From the question we are told that  

       The mass of the probe is  

       The location of the prob at time t = 22.9 s is  

       The  momentum at time  t = 22.9 s is  

        The net force on the probe is  

Generally the change in momentum is mathematically represented as

The initial time is   22.6 s

 The final time  is  22.9 s

             Substituting values  

Explanation

Rotational kinetic energy of the earth = 1/2 Iω²

where I is moment of inertia of the earth and ω is angular velocity .

I = 2/5 m R² , m is mass and R is radius of the earth .

I = 2/5 x 5.98 x 10²⁴ x ( 6.38 x 10⁶ )²

=97.36 x 10³⁶

ω = 1 / T

T = 24 x 60 x 60 = 86400 s

ω = 1 / 86400

= 11.57 x 10⁻⁶ rad / s

Rotational Kinetic energy = 1/2 Iω²

= .5 x 97.36 x 10³⁶ x (11.57 x 10⁻⁶ )²

= 6516.54 x 10²⁴ J