Variation in the Acceleration>Variation in Gravity with Altitude

The acceleration due to gravity, denoted by $g$, is the acceleration experienced by a body due to Earth's gravitational pull. Its value is generally calculated at the surface of the Earth. However, as an object moves to a greater height or altitude above the surface, its value changes. Specifically, the acceleration due to gravity decreases as the distance from the center of the Earth increases. Understanding this variation is crucial for fields like satellite mechanics and space travel.

The formulas for acceleration due to gravity (g) are provided below:

On the Earth's Surface:

At height $h$ above the Earth's Surface:

Simplified Formula for Small Altitudes ($h \ll R$):

Definition of Terms:

• g: Acceleration due to gravity on the Earth's surface.
• g_h: Acceleration due to gravity at height h above the Earth's surface.
• G: Universal Gravitational Constant.
• M: Mass of the Earth.
• R: Radius of the Earth.
• h: Altitude or height above the Earth's surface.

• The acceleration due to gravity (g_h) decreases as the altitude (h) of the body from the surface of the Earth increases.

• The simplified formula g_h = g (1 - \[\frac {2h}{R}\]) is valid only for small altitudes, where h << R.

The variation in g with altitude can be derived by comparing the gravitational acceleration at the surface (g) with the gravitational acceleration at height h (g_h).

Derivation Steps

1. Acceleration at the surface (g):
   \[g = \frac{G M}{R^2}\]
2. Acceleration at height h (g_h): At height h, the distance from the center of the Earth becomes (R + h).
   \[g_h = \frac{G M}{(R+h)^2}\]
3. Ratio of g_h to g: Divide the equation for g_h by the equation for g.
   \[\frac{g_h}{g} = \frac{\frac{G M}{(R+h)^2}}{\frac{G M}{R^2}}\]
4. Simplifying the Ratio: Cancel out the common terms (GM).
   \[\frac{g_h}{g} = \frac{R^2}{(R+h)^2}\]
5. Final Formula: Rearrange to solve for g_h.
   \[g_h = g \frac{R^2}{(R+h)^2}\]
   This equation clearly shows that g_h is proportional to R²/(R+h)², which is less than 1, confirming that g_h decreases with altitude.
6. Binomial Approximation for Small Altitude (h << R): The equation is rewritten by factoring out R² from the denominator:
   \[g_h = g \frac{R^2}{R^2 \left(1 + \frac{h}{R}\right)^2}\]
   \[g_h = g \left(1 + \frac{h}{R}\right)^{-2}\]
   For small altitude h, where \[\frac {h}{R}\] << 1, the Binomial Theorem approximation (1 + x)ⁿ ≈ 1 + nx is used (by neglecting higher power terms). Here x = h/R and n = -2.
   \[g_h \approx g \left(1 + (-2) \frac{h}{R}\right)\]
7. Simplified Formula:
   \[g_h = g \left(1 - \frac{2h}{R}\right)\]
   This simpler formula is used when the altitude h is much smaller than the radius R.

Prolem: At what distance above the surface of Earth the acceleration due to gravity decreases by 10% of its value at the surface? (Radius of Earth = 6400 km). Assume the distance above the surface to be small compared to the radius of the Earth.

Solution:

1. Determine the final value of g_h:

• g decreases by 10%, so g_h is 90% of g.
• \[\frac{g_h}{g} = \frac{90}{100} = 0.9\]

2. Use the simplified formula: Since the distance is assumed to be small, use the approximation:

3. Substitute the ratio and solve for h:

• 0.9 = 1 - \[\frac {2h}{R}\]
• h = \[\frac {0.1R}{2}\] = \[\frac {R}{20}\]

4. Calculate the value of h: Use the given value for the Radius of Earth (R = 6400 km).

• h = \[\frac {6400km}{20}\]
• h = 320 km

• Commercial Flights: At 10 km high, gravity is a bit weaker than on the ground.
• Mountain Climbing: Gravity is slightly less at the top of tall mountains like Everest.
• Low Earth Orbit (LEO) Satellites: In orbit (like the ISS), gravity is still strong (about 90%), but astronauts feel weightless because they’re always falling around Earth.