#### Topic Content:

- Gravitational Field Intensity
- Variation of g with Distance (r)
- Relationship between G and g
- Variation of g with Height

Gravitational field strength is the gravitational forceThe force of gravity, or gravitational force, pulls objects with mass toward each other. More 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.

Gravitational field strength, g, is a vector quantityVector quantities are quantities with magnitude and direction. Examples of vector quantities include displacement, velocity, position, force, and torque. More; The direction of g is always towards the **centre** of the body creating the gravitational field.

### Variation of g with Distance (r):

The earth is a sphere of mass M_{e} and radius r_{e}. We can take the distance between the centres of mass of Earth and an object on its surface to be the radius of Earth, provided that its size is much less than the radius of Earth.

The attractive force of the earth of mass (M_{e}) on another mass (m) is related to distance (r_{e}) from the earth by:

F = \( \frac{GM_em} {r_{e}^{2}}\) ………… (1)

- M
_{e}is the mass of the Earth - m is the mass of the object on the earth’s surface
- r
_{e}(the earth’s radius) is the distance of the object on the earth’s surface to the centre of the earth - G is the gravitational constant

F = mg ………. (2)

This is the force of gravity on the mass (m) due to the earth, where g is the acceleration due to gravity.

The force of gravity pulls all objects “downward” toward the planet’s centre.

The weight of the body at the centre of the earth is zero. This is because the value of g is zero at the centre of the earth. Because there is no separation between the two bodies i.e. Earth and the body under consideration.

Combining (1) and (2)

mg = \( \frac{GM_em} {r_{e}^{2}}\) ……….. (3)

cross multiply

g = \( \frac{GM_em} {r_{e}^{2}m}\)

g = \( \frac{GM_e \not{m}} {r_{e}^{2} \not{m}}\)

g = \( \frac{GM_e} {r_{e}^{2}}\) ………… (4)

This explains why all masses free fall with the same acceleration. We have ignored the fact that Earth also accelerates toward the falling object, but that is acceptable as long as the mass of Earth is much larger than that of the object.

Equation four (4) shows that:

- g is inversely proportional to the square of their distance apart
- Inverse square law is obeyed because as g decreases the distance from the earth increases

According to the equation, we should expect g to be slightly less on the tops of mountains than at sea level, since r_{e} is slightly greater at the top of the mountain than at sea level.

Also, the distance of poles, from the centre of the earth, is smaller than the distance of any other point on the earth’s surface from its centre. At the poles, no centrifugal force acts on the body. Hence the value of g is largest at the poles.

If we have accurate values for ‘g’ and ‘G’ we can use equation 4 to calculate the mass of the earth, M_{e}

M_{e} = \( \frac{gr_e^2}{G} \)

### Example 3.3.1:

Estimate the mass of the earth, given, the radius of the earth = 6.4 × 10^{6} m, acceleration due to gravity (g) = 9.8 m/s^{2} and gravitational constant (G) = 6.67 × 10^{−11} Nm^{2} kg^{-2}.

**Solution**

- M
_{e}= ? - r
_{e}= 6.4 × 10^{6}m - g = 9.8 m/s
^{2} - G = 6.67 × 10
^{−11}Nm^{2}kg^{-2}

M_{e} = \( \frac{gr_e^2}{G} \)

M_{e} = \( \frac{9.8 \: \times \: \left( 6.4 \: \times \: 10^6 \right)^2 }{6.67 \: \times \: 10^{-11}} \)

M_{e} = \( \frac{4.01 \: \times \: 10^{14} }{6.67 \: \times \: 10^{-11}} \)

M_{e} = \(\scriptsize 6 \: \times \: 10^{24} \: kg\)

### Variation of g with Height:

As the mass moves away from the centre of the earth, g also decreases

i.e. g = \( \frac{GM} {r^2}\) ……… (1)

At a height h from the earth’s surface,

g_{h} = \( \frac{GM}{r + h^2} = \frac{GM}{R^2}\) ………. (2)

where R = r + h

Dividing (2) by (1)

\( \frac{g_h}{g} = \frac{\frac{GM}{R^2}}{\frac {GM} {r^2}} \) \( \frac{g_h}{g} = \frac{GM}{R^2} \: \times \: \frac {r^2}{GM} \) \( \frac{g_h}{g} = \frac{r^2}{R^2} \) \( \frac{g_h}{g} = \left(\normalsize \frac{r}{R} \right)^2 \) \( \scriptsize g_h = g \left(\normalsize \frac{r}{R} \right)^2 \)### Example 3.3.2:

The acceleration due to gravity near the earth’s surface is about 10 ms^{-2}. Calculate the gravitational field strength at a height (h) twice the radius of the earth.

### Example 3.3.3:

What is the value of *g* 400 km above Earth’s surface, where the International Space Station is in orbit?

The radius of the earth = 6.37 × 10^{6} m

Mass of Earth = \(\scriptsize 5.96 \: \times \: 10^{24} \: kg\)

Gravitational constant (G) = 6.67 × 10^{−11} Nm^{2} kg^{-2}

**Solution**

g = \( \frac{GM_e} {R^{2}}\)

where R = r_{e} + 400 km

or R = r_{e} + 400 × 10^{3} m

g = \( \frac{6.67 \: \times \: 10^{-11} \: \times \: 5.96\:\times\:10^{24}} {6.37 \: \times \: 10^6 \: + \: 400 \: \times \: 10^3}\)

g = 8.67 m/s^{2}

### Example 3.3.4:

The value of acceleration due to gravity at a certain height h above the surface of the earth is \( \frac{g}{4}\), where g is the value of acceleration due to gravity at the surface of the earth. Calculate the height h.

(r is the radius of the earth)

**Solution**

where R = r + h

\( \frac{g}{4} = \scriptsize g \left(\normalsize \frac{r}{r\:+\:h} \right)^2 \) \( \frac{g}{4g} = \left(\normalsize \frac{r}{r\:+\:h} \right)^2 \) \( \frac{\not{g}}{4\not{g}} = \left(\normalsize \frac{r}{r\:+\:h} \right)^2 \) \( \frac{1}{4} = \left(\normalsize \frac{r}{r\:+\:h} \right)^2 \)take the square root of both sides

\( \sqrt{\frac{1}{4}} = \sqrt{\left(\normalsize \frac{r}{r\:+\:h} \right)^2} \) \( \frac{1}{2} = \left(\normalsize \frac{r}{r\:+\:h} \right) \)r + h = 2r

h = 2r – r

h = r