Question

1. The radius of the planet mars is approximately 3,396 km and has a gravitational acceleration equal to three eighths times than that of the earth. What is the velocity of escape? Ans. 5 km/s
2. Initially a tank holds 100 gal of a brine solution containing 20lb of salt. At t=0, fresh water is poured into the tank at the rate of 5 gal/min, while stirred mixture leaves the tank at the rate 3 gal/min. what is the amount of salt in the tank after 1 hours? How long will it take for the take to contain 10 lb of salt? Ans. 6.12 lb; 29.37 min.

Answer 1

1.

$$g_E=\dfrac{GM_E}{R_E^2}$$

$$g_{pl}=\dfrac{GM_{pl}}{R_{pl}^2}=\dfrac{3}{8}g_E$$

The velocity of escape is

$$v_{escape}=\sqrt{\dfrac{2GM_{pl}}{R_{pl}}}=\sqrt{\dfrac{2(\dfrac{3}{8}g_E)(R_{pl}^2)}{R_{pl}}}$$

$$v_{escape}=\sqrt{\dfrac{3g_ER_{pl}}{4}}$$

$$v_{escape}=\sqrt{\dfrac{3(9.81)(3396000)}{4}}=4998.6(m/s)$$

$5\ km/s$

2.

Let $s(t) =$ amount, in lb of salt at time $t.$ Then we have

$\dfrac{ds}{dt}=$(rate of salt into tank) − (rate of salt out of tank)

$$\dfrac{ds}{dt}=0-\dfrac{3s}{100+(5-3)t}$$

So we get the differential equation

$$\dfrac{ds}{dt}=-\dfrac{3s}{100+2t}$$

$$\int \dfrac{ds}{s}=-\int\dfrac{3dt}{100+2t}$$

$$s(t)=c_1(100+2t)^{-3/2}$$

$$s(0)=c_1(100)^{-3/2}=20$$

$$c_1=20000$$

$$s(t)=20000(100+2t)^{-3/2}$$

$$s(60)=20000(100+2(60))^{-3/2}$$

$$s(60)=6.13\ lb$$

$$s(t)=20000(100+2t)^{-3/2}=10$$

$$(100+2t)^{3/2}=2000$$

$$100+2t=100\sqrt[3]{4}$$

$$t=50(\sqrt[3]{4}-1)$$

$$t=29.37\ min$$