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Savatey [412]
3 years ago
14

At what altitude above the earth's surface would the acceleration due to gravity be 4.9 m/s^2? Assume the mean radius of the ear

th is 6.4x10^6 metres and the acceleration due to gravity 9.8m/s^2 on the surface of the earth.
Physics
1 answer:
Mandarinka [93]3 years ago
4 0
We apply the gravity calculation expressed in the formula: g=GM/r2 
where G is the gravitational constant, m is the mass and r is the radius
r=√GM/g
(1)       Radius = √6.674e-11*5.972e24/8             = 7058 kms    Earth radius or surface of earth from center of earth= 6400 kmsSo r= 658 kms from surface of earth.
Gravity 8m/s2 will be at 658 kms from surface of earth.
(2) half gravity= 9.8/2= 4.9 m/s2     Radius=√6.674e-11*5.972e24/4.9                 = 9019 kms        Half Gravity will exist at 9019-6400= 2619 kms from surface of earth.
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Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

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Let u is the object distance. It can be calculated using the mirror's formula as :

\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}

\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}

\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}

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The magnification of the mirror is given by :

m=\dfrac{-v}{u}

m=\dfrac{-(-37.5)}{(-21.09)}

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