At what height from the surface of earth the gravitation potential and the value of g are -5.4 × 10^7 J kg^-1

Question

At what height from the surface of earth the gravitation potential and the value of g are -5.4 × 10^7 J kg^-1 and 6.0 m s^-2 respectively? (Take the radius of earth as 6400 km.)

Accepted Answer

Let $r$ be the distance from the center of the Earth to the point at height $h$. 
The gravitational potential $V$ at distance $r$ is given by magnitude $|V| = \frac{GM}{r}$. 
The acceleration due to gravity $g_h$ at distance $r$ is given by $g_h = \frac{GM}{r^2}$. 
Dividing the magnitude of potential by the acceleration gives the distance $r$: 
$\frac{|V|}{g_h} = \frac{GM/r}{GM/r^2} = r$. 
Substituting the given values: 
$r = \frac{5.4 	imes 10^7 	ext{ J/kg}}{6.0 	ext{ m/s}^2} = 0.9 	imes 10^7 	ext{ m} = 9 	imes 10^6 	ext{ m} = 9000 	ext{ km}$. 
The height $h$ from the surface is $h = r - R$, where $R$ is the Earth's radius (6400 km). 
$h = 9000 	ext{ km} - 6400 	ext{ km} = 2600 	ext{ km}$.

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