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Lesson 3, Topic 3
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Gravitational Field Intensity

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Gravitational field strength is the gravitational force per unit mass at a particular point. Gravitational field strength

G =$$\frac{Gravitational \; force}{mass} = \frac{F}{m}$$

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)

$$\frac{g^1}{g} = \frac{GM}{(r + h^2)} = \frac{r^2}{R^2}$$  (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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