Derivation: Acceleration Due to Gravity with Altitude ( Above or Height ) – Formula, Derivation, FAQs , MCQs
Variation of Acceleration Due to Gravity with Altitude (Height))
Acceleration due to gravity (g) decreases as we move away from the Earth's surface. This variation can be explained using Newton's law of gravitation.
Acceleration Due to Gravity on the Surface of Earth
The acceleration due to gravity at the Earth's surface is given by:
$$g=\frac{GM}{R^2}$$
Where:
- G = Universal Gravitational Constant
- M = Mass of the Earth
- R = Radius of the Earth
Acceleration Due to Gravity at Height h
Consider a body at a height h above the Earth's surface.
Distance from the centre of Earth = (R + h)
Therefore, acceleration due to gravity at height h is:
$$g_h=\frac{GM}{(R+h)^2}$$
Derivation of Relation Between g and gh
Dividing the above equation by the surface gravity equation:
$$\frac{g_h}{g}=\frac{\frac{GM}{(R+h)^2}}{\frac{GM}{R^2}}$$
$$\frac{g_h}{g}=\frac{R^2}{(R+h)^2}$$
$$\frac{g_h}{g}=\frac{1}{\left(1+\frac{h}{R}\right)^2}$$
Hence,
$$g_h=\frac{g}{\left(1+\frac{h}{R}\right)^2}$$
or
$$g_h=g\left(1+\frac{h}{R}\right)^{-2}$$
Approximate Formula for Small Heights
If h << R, then using Binomial Theorem:
$$\left(1+\frac{h}{R}\right)^{-2}\approx\left(1-\frac{2h}{R}\right)$$
Therefore,
$$g_h=g\left(1-\frac{2h}{R}\right)$$
Decrease in the Value of g
The decrease in acceleration due to gravity is:
$$g-g_h=\frac{2gh}{R}$$
Percentage Decrease in g
$$\%\text{ Decrease}=\frac{g-g_h}{g}\times100$$
$$\%\text{ Decrease}=\frac{2h}{R}\times100$$
Important Facts
- Acceleration due to gravity decreases as altitude increases.
- For large heights (comparable to Earth's radius), use:
$$g_h=\frac{g}{\left(1+\frac{h}{R}\right)^2}$$
- For small heights, use:
$$g_h=g\left(1-\frac{2h}{R}\right)$$
- The weight of a body decreases with height because g decreases.
Weight of a body:
$$W=mg$$
Conclusion
The acceleration due to gravity decreases with increasing altitude because the distance between the body and the Earth's centre increases. For small heights, the decrease is approximately linear, whereas for large heights the exact formula must be used.
Frequently Asked Questions (FAQs)
1. Why does acceleration due to gravity decrease with height?
As the distance from the Earth's centre increases, gravitational attraction decreases, causing g to decrease.
2. What is the exact formula for gravity at height h?
$$g_h=\frac{g}{\left(1+\frac{h}{R}\right)^2}$$
3. What is the approximate formula for small heights?
$$g_h=g\left(1-\frac{2h}{R}\right)$$
4. Does weight decrease with altitude?
Yes. Since W = mg, a decrease in g causes a decrease in weight.
5. Is gravity zero in space?
No. Gravity decreases with distance but never becomes exactly zero.
MCQ Quiz
1. The acceleration due to gravity at the Earth's surface is:
A) $$g=\frac{GM}{R}$$
B) $$g=\frac{GM}{R^2}$$
C) $$g=\frac{GM^2}{R^2}$$
D) $$g=\frac{GR}{M}$$
Answer: B
2. The acceleration due to gravity at a height h above the Earth's surface is:
A) $$\frac{GM}{R+h}$$
B) $$\frac{GM}{R^2}$$
C) $$\frac{GM}{(R+h)^2}$$
D) $$\frac{GM}{(R-h)^2}$$
Answer: C
3. The exact relation between gh and g is:
A) $$g\left(1+\frac{h}{R}\right)^2$$
B) $$\frac{g}{\left(1+\frac{h}{R}\right)^2}$$
C) $$g\left(1-\frac{h}{R}\right)$$
D) $$\frac{g}{1+\frac{h}{R}}$$
Answer: B
4. For h << R, the approximate expression for gh is:
A) $$g\left(1+\frac{2h}{R}\right)$$
B) $$g\left(1-\frac{h}{R}\right)$$
C) $$g\left(1-\frac{2h}{R}\right)$$
D) $$g\left(1+\frac{h}{R}\right)$$
Answer: C
5. As altitude increases, the value of g:
A) Increases
B) Remains constant
C) First increases then decreases
D) Decreases
Answer: D
Yes
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