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JAMB Physics 2019 Paper

Question : 34

Total: 100

^{-11}

^2

^2

^{24}

^6

^2

)

Solution:

Concept:

Gravitational field intensity at the surface is the gravitational force acting on a unit mass.

It is numerically equal to acceleration due to gravity, so

E = g

Explanation:

For a mass

m

on the earth’s surface, the gravitational force is

F = \frac{GMm}{R^2}

Field intensity is

E = \frac{F}{m} = \frac{GM}{R^2}

Substitute

G = 6.7 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2

M = 6 \times 10^{24} \, \text{kg}

, and

R = 6.4 \times 10^6 \, \text{m}

E = \frac{6.7 \times 10^{-11} \times 6 \times 10^{24}}{(6.4 \times 10^6)^2}

E = \frac{4.02 \times 10^{14}}{4.096 \times 10^{13}} = 9.8 \, \text{N/kg}

Since

1 \, \text{N/kg} = 1 \, \text{m/s}^2

, this equals the known value

g = 9.8 \, \text{m/s}^2

Answer:

9.8 \, \text{N/kg}

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