Question

The value of $g$ at the surface of earth is $9.8{\text{m/s}}^2$. Then the value of $g^{\prime }$ at a place $480\text{km}$ above the surface of the earth will nearly be (radius of the earth is $\text{6400}\text{km}$)

• A. $9.8{\text{m/s}}^2$
  
• B. $7.2{\text{m/s}}^2$
  
• C. $8.5{\text{m/s}}^2$
  
• D. $4.2{\text{m/s}}^2$
  

Answer 1

The correct option is C $8.5{\text{m/s}}^2$


Acceleration due to gravity at a height $h$ above the surface of the Earth, $$g_h=\frac{GM}{(R+h{)}^2}..\left(1\right)$$
But accelaration due to gravity at the surface of the Earth,$$g=\frac{GM}{R^2}..\left(2\right)$$ From equation $\left(1\right)$ and $\left(2\right)$, we get

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

Substituting, $R=6400\text{km},h=480\text{km}$ and $g=9.8{\text{m/s}}^2$

$\Rightarrow g_h=9.8\times {\left[\frac{6400}{6400+480}\right]}^2=8.48{\text{m/s}}^2\approx 8.5{\text{m/s}}^2$


Hence, option (c) is the correct answer.

$\text{Key Concept :}$ The acceleration due to gravity due to a planet or any point mass is $${\frac{GM}{r}}^2$$ $\text{Why this question:}$ The result $g^{\prime }=g(1-2h/Re)$ can only be used when $h<<R$.