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ziro4ka [17]
3 years ago
11

How far (in meters) above the earth's surface will the acceleration of gravity be 21.0 % of what it is on the surface?

Physics
1 answer:
weeeeeb [17]3 years ago
8 0

Answer:

7532m

Explanation:

Gravity on the surface of the earth with radius R is given by:

F_1=\frac{Gm_{earth}m_{object}}{R^2_{earth}}

Gravity a distance r above the surface:

F_2=\frac{Gm_{earth}m_{object}}{(R+r)^2}

How big is r if:

F_2=0.21\times F_1

You get the following equation:

R^2=0.21(R+r)^2=0.21R^2+0.41Rr+0.21r^2

Solve for r:

r^2+2Rr-\frac{79}{21} R^2=0

r=-R+\sqrt{R^2+\frac{79}{21}R^2}=-R+\sqrt{\frac{100}{21}R^2}=(\frac{10}{\sqrt{21}}-1)R

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8 0
3 years ago
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AleksAgata [21]
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5 0
3 years ago
The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give
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Acceleration is given by the formula ;

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8 0
3 years ago
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