AT WHAT DISTANCE ABOVE THE SURFACE OF THE EARTH IS THE ACCELERATION DUE TO EARTH'S GRAVITY 0.980M/S² IF THE ACCELERATION DUE TO GRAVITY AT THE SURFACE HAS MAGNITUDE 9.80M/S²

Well, if you're searching for that specific distance above the surface of the Earth where the acceleration due to Earth's gravity is 0.980 m/s², I have to warn you - it's a high-flying task! But fear not, for the answer is no laughing matter.

To tackle this question, let's think about how gravity behaves. As we move further away from the surface of the Earth, the acceleration due to gravity gradually decreases. We're looking for the point where it reaches 0.980 m/s², right?

So, if we start with a value of 9.80 m/s² at the Earth's surface, we can divide this value by 10 to get a rough estimate. This tells us that for every 10 meters we go up, the acceleration due to gravity decreases by about 1 m/s².

Using this calculation, we find that at around 98 meters above the Earth's surface, the acceleration due to gravity drops to 0.980 m/s². At this height, you'll experience a force of gravity that's approximately one-tenth of what you feel on the surface.

Remember, though, this estimation assumes a constant decrease in gravity with distance, which is not entirely accurate. Gravity's behavior can get a bit wonky as we move further away, but it should provide a reasonable ballpark estimation.

I hope that answers your gravity-related query without being too deflating!

To find the distance above the surface of the Earth where the acceleration due to gravity is 0.980 m/s², we can use the formula for the acceleration due to gravity at different distances from the Earth's center.

The formula for the acceleration due to gravity at a certain distance (h) above the Earth's surface is given by:

g(h) = g₀ / (1 + h / R)²

Where:
g(h) is the acceleration due to gravity at height h above the Earth's surface,
g₀ is the acceleration due to gravity at the Earth's surface (9.80 m/s²),
R is the radius of the Earth (approximately 6,371,000 meters).

We need to solve the equation for g(h) = 0.980 m/s².

0.980 = 9.80 / (1 + h / 6,371,000)²

Now, let's solve this equation step-by-step.

Step 1: Rearrange the equation:
(1 + h / 6,371,000)² = 9.80 / 0.980

Step 2: Simplify the equation:
(1 + h / 6,371,000)² = 10

Step 3: Take the square root of both sides:
1 + h / 6,371,000 = √10

Step 4: Subtract 1 from both sides:
h / 6,371,000 = √10 - 1

Step 5: Multiply both sides by 6,371,000:
h = 6,371,000 * (√10 - 1)

Step 6: Calculate the value:
h ≈ 5,790,100 meters

Therefore, the distance above the surface of the Earth where the acceleration due to gravity is 0.980 m/s² is approximately 5,790,100 meters.

To find the distance above the surface of the Earth where the acceleration due to gravity is 0.980 m/s², we can use the inverse square law of gravity.

The inverse square law states that the gravitational force between two objects is inversely proportional to the square of the distance between their centers. Mathematically, it can be expressed as:

F = (G * m₁ * m₂) / r²

Where:
F is the gravitational force
G is the gravitational constant (approximately 6.674 × 10⁻¹¹ N m²/kg²)
m₁ and m₂ are the masses of the objects
r is the distance between the centers of the objects

In this case, we know that the acceleration due to gravity at the Earth's surface is 9.80 m/s². This can be used to find the mass of the Earth (m₂) using Newton's second law:

F = m₂ * a

Where a is the acceleration due to gravity at the Earth's surface. Rearranging the equation, we have:

m₂ = F / a

m₂ = (G * m₁ * m₂) / r² / a

m₂ = G * m₁ / (r² * a)

Now, we can find the distance (r) at which the acceleration due to gravity is 0.980 m/s². Rearranging the equation again, we have:

r² = G * m₁ / (m₂ * a)

r = √(G * m₁ / (m₂ * a))

Plugging in the known values:
G = 6.674 × 10⁻¹¹ N m²/kg²
m₁ = mass of the Earth (approximately 5.972 × 10²⁴ kg)
a = 9.80 m/s²
m₂ = mass of an object (which we are trying to find)

r = √((6.674 × 10⁻¹¹ N m²/kg² * 5.972 × 10²⁴ kg) / (m₂ * 9.80 m/s²))

Solving this equation will give us the distance (r) above the surface of the Earth where the acceleration due to gravity is 0.980 m/s².

radius of earth = 6378.1 km

Assuming g is not affected by other heavenly bodies, then Newton's law of gravitation applies:
g=GMm/r^2=k/r^2 (k=a constant)
=>
g'r'^2=gr^2
r'^2=r^2(g/g')
r'=r sqrt(g/g')
=6378.1 sqrt(9.80/0.980)
= 2.02*10^4 km approx.