Solved Examples on Variation in Acceleration due to Gravity
Example 1: A man, Ravi, weighs on the surface of the earth. At what height will Ravi’s weight become half, which is reduced to W/2? Assume the radius of the earth to be R.
Solution:
Given that,
Wh = W / 2
Radius of the Earth = R
We have to find R,
Thus, r = R + h
or,
h = r – R
= 1.4.14R – R
= 0.414R
Therefore, at the height of 0.414R, a man’s weight becomes half his weight on the surface of the earth.
Example 2: The earth is assumed to be spherically homogeneous for all practical purposes. Find the acceleration due to gravity at a depth of 2000 km beneath the surface of the earth. [Use: R = 6400 km and g = 9.8 m/s2.]
Solution:
Given that:
The depth, d = 2000 km,
The radius of the earth, R = 6400 km.
We have to find acceleration due to gravity gd
Therefore, the acceleration at a depth of 2000 km below the surface of the earth is 6,738 m/s2.
Example 3: Point P is located at the surface of the earth, with a radius of 6400 km. The particle accelerates downward with a constant acceleration to reach a point Q situated at a depth d equal to 1600 km from the earth’s surface. Find the reduction witnessed in the acceleration due to gravity at Q.
Solution:
Given that,
The depth, d = 1600 km,
The radius of the earth, R = 6400 km,
Also the acceleration due to gravity, g = 9.8 m/s2
We have to find the decrease in value of acceleration due to gravity i.e. g – gd
Hence,
g – gd = 9.8 m/s2 – 7.35 m/s2
= 2.45 m/s2
Therefore, the decrease in acceleration due to gravity is 2.45 m/s2.
Example 4: A person weighs 50 kg on Earth. On the surface of Neptune, where the acceleration of gravity is approximately 1.2 times as great as that on Earth, what will be the person’s mass?
Solution:
The person will weigh 50 kg on the surface of Neptune because the force of gravity changes the weight of the object and not it’s mass. The mass of an object stays constant regardless of the forces acting upon it.
Example 5: It has been noted that at a place above the earth’s surface, the value of g is 0.2 m/s2. Determine this height above the earth’s surface at which the given g is obtained.
Solution:
Consider the height above earth surface be h’.
Therefore,
g′ = g (1− 2h / R)
or
g′ / g = (1− 2h / R)
Substitute the given values in the above expression as,
0.2 m/s2 / 9.8 m/s2 = (1− 2h / R)
2h / R = 48 / 49
h = 3134. 7 km
= 3134.7 km
Variation in Acceleration due to Gravity
The value of gravity’s attraction or potential is governed by the distribution of mass within Earth or another celestial body. As noted previously, the distribution of matter affects the geometry of the surface where the potential is constant. Gravity and potential measurements are thus critical to both geodesy, which studies the form of the Earth, and geophysics, which studies its interior structure. The orbits of artificial satellites are the greatest way to assess the potential for play an intermediary role and global geophysics. Surface gravity measurements are ideal for local geophysics, which studies the structure of mountains and seas as well as the hunt for minerals. In order to understand how the value of acceleration due to gravity affected lets first under its basics as:
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