Question

1. A certain planet has an average surface area that is 2.5% larger than the earth and a gravitational acceleration that is one fifth times than that of the earth’s gravitational acceleration. What is the velocity of escape on that planet? Ans. 5.04 km/s
2. An object weighing 16 pounds is dropped from the top of a 1000 ft building with an initial velocity of 10 ft/sec. as it falls, the object encounters air resistance that is equal to 1/4 v. What is the velocity of the object at the end of 5 seconds? What is the limiting velocity? Ans. 59.57 ft/s; 64 ft/s

Answer 1

Solution;

(1)

$g_p=\frac15×9.81=1.962m/s^2$

Calculate surface are of earth;

$S.A_e=4πr^2$

$S.A_e=4×π×6400^2=514718540.4km^2$

Now;

$S.A_p=1.025S.A_e=527586503.9km^2=4πr_p^2$

Hence;

$r_p=6479.51km$

But;

$g_p=\frac{GM}{r_p^2}$ My

Therefore;

$M_p=\frac{g_p×r_p^2}{G}=\frac{1.962×(6479510)^2}{6.674×10^{-11}}$

$M_p=1.234×10^{24}kg$

The escape velocity;

$v_e=\sqrt{\frac{2GM}{r}}$

$v_e=\sqrt{\frac{2×6.674×10^{-11}×1.234×10^{24}}{6479510}}$

$v_e=5042.37m/s$

$v_e=5.04km/s$

(2)

W=16lb

W=mg

$m=\frac Wg=\frac{16}{32}=0.5lbm$

$V(0)=10ft/s$

Aire resistance =0.25v

Hence;

$\frac{dv}{dt}+0.5v=32$

Integrating factor;

$I.F=e^{0.5dt}=e^{0.5t}$

The solution is;

$ve^{0.5t}=32\int e^{0.5t}dt+C$

$v(t)=\frac{64e^{0.5t}+C}{e^{0.5t}}$

$v(t)=64+Ce^{0.5t}$

$v(0)=10ft/s$

$10=64+C$

Therefore;

$C=-54$

Hence;

$v(t)=64-54e^{0.5t}$

Limiting velocity is ;

$64ft/s$

Velocity after 5 second;

$v(5)=64-54e^{0.5×5}=-59.57ft/s$