A satellite orbits the Earth at an altitude equal to three times the radius of the Earth, … — JAMB UTME

A satellite orbits the Earth at an altitude equal to three times the radius of the Earth, $R$ . If the acceleration due to gravity at the Earth's surface is $g$ , what is the gravitational field strength experienced by the satellite at its orbit?

$\frac{g}{16}$ Answer
B
$\frac{g}{9}$
C
$\frac{g}{4}$
D
$\frac{g}{3}$

Answer

The gravitational field strength at the orbital position is

\frac{g}{16}

The total distance from the center of the Earth to the satellite is

r = R + 3R = 4R

. Since gravitational field strength is inversely proportional to the square of the distance from the center of mass, increasing the distance by a factor of

4

reduces the field strength by a factor of

4^2 = 16

, giving

\frac{g}{16}

Step-by-Step Solution

Determine the total distance from the center of the Earth to the satellite.

The orbital radius is

r = R + h = R + 3R = 4R

Gravitational calculations must be measured from the center of mass of the attracting body, not its surface.

Apply Newton's law of universal gravitation for gravitational field strength.

At the surface (

r = R

g = \frac{GM}{R^2}

. At orbit (

r = 4R

g' = \frac{GM}{(4R)^2} = \frac{GM}{16R^2}

Gravitational field strength follows an inverse-square law with respect to distance from the center of mass.

Express the orbital gravitational field strength in terms of the surface gravity

g

g' = \frac{1}{16} \left(\frac{GM}{R^2}\right) = \frac{g}{16}

Substituting the surface value

g = \frac{GM}{R^2}

yields the simplified ratio.

Key Concept

Gravitational Field Strength and Inverse-Square Law

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