Gravitational field strength is the gravitational force per unit mass at a particular point. Gravitational field strength
G =\( \frac{Gravitational \; force}{mass} =Â \frac{F}{m}\)
F = mg
The gravitational field strength (g) is also known as acceleration due to gravity. The unit of g is N/kg and is constant at a particular location.
Variation of g with Distance (r)
The attractive force of the earth on another mass (m) is related to distance (r) from the earth by:
 F = \( \frac{GMM} {r^2}\) ________ (1)
 F=mg _______ (2)
Combining (1) and (2)
 Mg = \( \frac{GMM} {r^2}\)  ___________ (3)
 g = \( \frac{GM} {r^2}\) __________ (4)
This equation four shows that
- g is inversely proportional to the square of their distance apart
- Varies as the radius of the earth
- Obeys inverse square law as g decreases as mass distance from the earth increases
Variation of g with Height
As the mass moves away from the center of the earth, g also decreases
 I.e. g = \( \frac{GM} {r^2}\) _________ (1)
At a height h from the earth surface,
 g=\( \frac{GM}{r + h^2} = \frac{GM}{R^2}\) (2)
Dividing (1) by (2)
\( \frac{g^1}{g} =Â \frac{GM}{(r + h^2)} = \frac{r^2}{R^2}\)Â (2)
\( \scriptsize g^1 = g \left(\normalsize \frac{r}{R} \right)^2 \)Example
The acceleration due to gravity near the earth surface is about10ms-2. Calculate the gravitational field strength at a height (h) twice the radius of the earth.
\( \scriptsize g^1 = g \left(\normalsize \frac{r}{R} \right)^2 \) \( \scriptsize g^1 = 10 \left(\normalsize \frac{r}{R} \right)^2 \) \( \scriptsize g^1 = 10 \left(\normalsize \frac{1}{2} \right)^2 \)= 2.5ms-2
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