Question

a satellite's escape velocity is 6.5 mi/sec, the radius of the earth is 3960 mi, and the earth's gravitational constant is 32.2 ft/sec^2. How far is the satellite from the surface of the earth

Answer 1

Answer:

$23.89x10^{6}$ ft

Step-by-step explanation:

$\frac{V^{2} }{R^{2} } = \frac{2g}{R+h}$ This is the formula to find satellite's escape velocity V , where R is earth's radius, h is the satellite's height from the earth surface and g is the earth's gravitational constant.

$R^{2}(R+h) \frac{V^{2} }{R^{2} } =R^{2} (R+h)\frac{2g}{R+h}$   (Multiplying to clear the fractions)

(R+h)$V^{2} =R^{2} 2g$

$R+h=\frac{2R^{2}g }{V^{2} }$

$h= \frac{2R^{2} g}{V^{2} } -R$

Now, we can determine height of satellite from the surface of the earth

by putting the values in above equation

$h= \frac{2 * (3960)^{2}*32.2 }{6.5^{2} } -3960\\h= 23.902 x10^{6} - 3960\\ h=23.89x10^{6} ft$