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If the acceleration due to gravity, g, is 10 m/s2 at the surface of earth (radius 6400 km), then at a height 1600 km the value of g (in m/s2) will be
1. 6.4
2. 5
3. 7.5
4. 2.5

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Correct Answer - Option 1 : 6.4

Concept:

Acceleration Due to Gravity:

  • The force of attraction exerted by the earth on a body is called gravitational pull or gravity.
  • We know that when a force acts on a body, it produces acceleration. Therefore, a body under the effect of gravitational pull must accelerate.
  • The acceleration produced in the motion of a body under the effect of gravity is called acceleration due to gravity, it is denoted by g.
  • Value of g changes as we go above and below the surface of earth.

∴ Acceleration due to gravity at some height (gh) is given by:

 \({g_h} = \frac{g}{{{{\left( {1 + \frac{h}{R}} \right)}^2}}} = g\left( {1 - \frac{{2h}}{R}} \right)\) this is true if h≪ R

where, G = universal gravitational constant, M = mass of the earth, R = radius of the earth, g = acceleration due to gravity on planet’s surface, gh = acceleration due to gravity at some height “h” from planet’s surface

Calculation:

Given:

g = 10 m/s2, h = 1600 km, R = 6400 km

Acceleration due to gravity at height h (gh) is:

\({g_h} = \frac{g}{{{{\left( {1 + \frac{h}{R}} \right)}^2}}}\)

\({g_h} = \frac{g}{{{{\left( {1 + \frac{1600}{6400}} \right)}^2}}}= \frac{16g}{25}\)

\({g_h} =\frac{16\times10}{25} =6.4~m/s^2\)

Variation in g with depth:

\(g' = g\left[ {1 - \frac{d}{R}} \right]\)

Where is the g = Acceleration due to gravity at depth d, R = Radius of the earth, and g = Acceleration due to gravity at the surface of the earth.

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