A 2.0kilogram mass is located 3.0 meters above the surface of Earth. What is the magnitude of Earth’s gravitational field strength at this location? (1) 4.9 N/
Question
the surface of Earth. What is the magnitude of
Earth’s gravitational field strength at this
location?
(1) 4.9 N/kg (3) 9.8 N/kg
(2) 2.0 N/kg (4) 20. N/kg
2 Answer
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1. User Answers PhyCS
(3) 9.8 N/kg is your answer. 
2. User Answers IsrarAwan
The correct answer is: Option (3) 9.8 N/kg
Explanation:
According to Newton's Law of Gravitation:
[tex] F_g = \frac{GmM}{R^2} [/tex]  (1)
Where G = Gravitational constant = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²
m = Mass of the body = 2 kg
M = Mass of the Earth = 5.972 × 10²⁴ kg
R = Distance of the object from the center of the Earth = Radius of the Earth + Object's distance from the surface of the Earth = (6371 * 10³) + 3.0 = 6371003 m
Plug in the values in (1):
(1)=> [tex] F_g = \frac{6.67408 * 10^{11} * 2 * 5.972*10^{24}}{(6371003)^2} = 19.63 [/tex]
Now that we have force strength at the location, we can use:
Force = mass * gravitationalfieldstrength
Plug in the values:
19.63 = 2.0 * gravitationalfieldstrength
gravitationalfieldstrength = 19.63/2 = 9.82 N/kg
Hence the correct answer is Option (3) 9.8 N/kg