Gravitational acceleration at the surface of the Earth is 9.8 m/s². What will be the approximate acceleration value from the Earth's surface at $${\frac{1}{{10}}^{th}}$$ the height of its radius?

Solution (By Examveda Team)

What is Gravitational Acceleration (g)?
Imagine dropping a ball. It falls faster and faster! This increase in speed is due to gravitational acceleration. On the surface of the Earth, this acceleration is about 9.8 m/s². This means for every second the ball falls, its speed increases by 9.8 meters per second.

Does Gravity Change with Height?
Yes! The force of gravity gets weaker the farther you are from the center of the Earth. So, if you go up into the sky, gravity won't pull you down quite as hard as it does on the ground. This means the gravitational acceleration will be less than 9.8 m/s².

The Formula We Use:
To find the new gravitational acceleration (let's call it g') at a certain height (h) above the Earth's surface, we use this special formula:
g' = g * (R / (R + h))²

Let's understand what each part means:
- g': This is the new gravitational acceleration we want to find at height 'h'.
- g: This is the gravitational acceleration on the Earth's surface, which is given as 9.8 m/s².
- R: This is the radius of the Earth (the distance from the Earth's center to its surface). We don't need its actual number because it will cancel out in our calculation.
- h: This is the height above the Earth's surface. The question tells us this height is 1/10th of the Earth's radius. So, we can write h = R/10.

Now, Let's Calculate Step-by-Step!
1. First, substitute the value of 'h' into our formula:
We know h = R/10.
So, our formula becomes: g' = 9.8 * (R / (R + R/10))²

2. Next, simplify the part inside the bracket (R + R/10):
Imagine you have one whole pizza (R) and you add one-tenth of a pizza (R/10). Together, you have 11/10 of a pizza!
So, R + R/10 = 10R/10 + R/10 = 11R/10

3. Put this simplified value back into the formula:
g' = 9.8 * (R / (11R/10))²

4. Simplify the fraction (R divided by 11R/10):
When you divide R by 11R/10, the 'R' on top and bottom cancels out. It becomes:
R * (10 / 11R) = 10/11

5. Now, square this fraction:
(10/11)² = (10 * 10) / (11 * 11) = 100 / 121

6. Finally, multiply this by 'g' (which is 9.8 m/s²):
g' = 9.8 * (100 / 121)
g' = 980 / 121
If you do this division, you get: g' ≈ 8.099 m/s²

7. Rounding this to one decimal place, we get approximately 8.1 m/s².

Does this Answer Make Sense?
Yes! We expected the gravitational acceleration to be less than 9.8 m/s² because we are at a height above the Earth's surface. Our calculated value of 8.1 m/s² is indeed less than 9.8 m/s², which is a good sign!

Comparing our result with the options:
Option A: 4.0 m/s²
Option B: 8.9 m/s²
Option C: 4.5 m/s²
Option D: 8.1 m/s²

Our calculated value perfectly matches Option D.

So, at 1/10th the height of Earth's radius, the gravitational acceleration will be approximately 8.1 m/s²!