Final answer:
The height of the meteor from the Earth's surface where its acceleration due to gravity becomes 4m/s² is calculated using Newton's law of gravitation and the known values of Earth's mass and radius. The result is a height of 3600 km.
Step-by-step explanation:
To find the height at which the acceleration due to gravity falls to 4 m/s2, we use Newton's law of gravitation. This law states that the force of gravity between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The formula is given by:
F = G (m1 * m2) / r2,
where F is the force of gravity, G is the gravitational constant (6.674 x 10-11 Nm2/kg2), m1 is the mass of the first object, m2 is the mass of the second object, and r is the distance between the centers of the two masses.
For an object near Earth, this force translates into the acceleration due to gravity using the formula:
g = G * M / r2
Where g is the acceleration due to gravity, M is the mass of the Earth, and r is the distance from the object to the center of the Earth, which is equal to the Earth's radius plus the height of the object above the Earth's surface.
Given the acceleration due to gravity (g) at a certain height is 4 m/s2, the mass of the Earth (M) is 6×1024 kg, and Earth's radius (R) is 6.4×103 km, we can rearrange the formula to solve for r.
g = G * M / r2
r = sqrt(G * M / g)
Plugging in the values and solving for r gives us the distance from the center of the Earth to the meteor. To find the height (h) from the Earth's surface, subtract Earth's radius from this distance:
h = r - R
After solving, we find that h = 3600 km, which is the height above Earth's surface where the acceleration due to gravity is 4 m/s2.