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Citrus2011 [14]
3 years ago
14

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
Mathematics
1 answer:
Artyom0805 [142]3 years ago
5 0

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

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