Answer:
the height of the meteor from the Earth's surface where its acceleration due to gravity becomes 4 m/s^2 is approximately 2.4 × 10^3 meters.
EXPANATION
To find the height of the meteor from the Earth's surface where its acceleration due to gravity becomes 4 m/s^2, we can use the formula for gravitational acceleration:
g = G * (M / r^2)
Where:
g is the acceleration due to gravity,
G is the gravitational constant (approximately 6.67 × 10^-11 Nm^2/kg^2),
M is the mass of the Earth,
and r is the distance from the center of the Earth.
First, let's find the acceleration due to gravity at the Earth's surface. We can plug in the given values:
g = G * (M / r^2)
g = (6.67 × 10^-11 Nm^2/kg^2) * (6 × 10^24 kg) / (6.4 × 10^3 m)^2
g ≈ 9.8 m/s^2
Since the acceleration due to gravity on the Earth's surface is approximately 9.8 m/s^2, we need to find the height where the acceleration becomes 4 m/s^2.
We can set up a proportion to find the height (h):
g / (r + h)^2 = 4 / r^2
Cross-multiplying, we get:
g * r^2 = 4 * (r + h)^2
Expanding the right side, we have:
g * r^2 = 4 * (r^2 + 2rh + h^2)
Simplifying further:
g * r^2 = 4r^2 + 8rh + 4h^2
Subtracting g * r^2 from both sides:
0 = 3r^2 + 8rh + 4h^2
Now, we can solve this quadratic equation for h. However, since the values of r and g are quite large, we can approximate this equation by ignoring the 4h^2 term, assuming it is relatively small compared to the other terms. This allows us to solve for h more easily.
0 ≈ 3r^2 + 8rh
Solving for h:
h ≈ -3r^2 / (8r)
Simplifying further:
h ≈ -3r / 8
Using the given values:
h ≈ -3 * (6.4 × 10^3 m) / 8
h ≈ -2.4 × 10^3 m
Since distance cannot be negative, we can take the absolute value of h to find the height:
h ≈ 2.4 × 10^3 m
Therefore, the height of the meteor from the Earth's surface where its acceleration due to gravity becomes 4 m/s^2 is approximately 2.4 × 10^3 meters.
