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Assuming that the Earth has a uniform density =5540.0 kg/m3,

what is the value of the gravitational acceleration
at a distance =900.0 km
from the Earth's center?

User Jcs
by
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1 Answer

7 votes

Answer:

8.957 m/s^2

Step-by-step explanation:

First, let's calculate the mass of the Earth using the formula:

M = (4/3) * π * R^3 * ρ

where:

- π is approximately 3.14159

- R is the radius of the Earth in meters, which is approximately 6,371,000 m

- ρ is the density of the Earth in kg/m^3, which is given as 5540.0 kg/m^3

M = (4/3) * 3.14159 * (6,371,000)^3 * 5540.0

M ≈ 5.972 × 10^24 kg

Now, we can use the formula for gravitational acceleration:

g = (G * M) / r^2

where:

- G is the gravitational constant, which is approximately 6.67430 × 10^-11 m^3/(kg * s^2)

- M is the mass of the Earth, which we calculated as 5.972 × 10^24 kg

- r is the distance from the Earth's center in meters, which is 900,000 m

Plugging in the values, we get:

g = (6.67430 × 10^-11 * 5.972 × 10^24) / (900,000)^2

g ≈ 8.957 m/s^2

Therefore, at a distance of 900.0 km from the Earth's center, the value of gravitational acceleration is approximately 8.957 m/s^2.

User Kyuuuyki
by
8.7k points