What is the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is 1/45 of its value at the Earth's surface?
45^2=r^2/r^2....^2cancel's. Take the square root of 45*earths radius 6,380,000(in meters)=4.27x10^7
To find the distance from the Earth's center to a point outside the Earth, where the gravitational acceleration is 1/45 of its value at the Earth's surface, we can use the concept of gravitational acceleration and the inverse square law.
First, we need to understand that the gravitational acceleration due to an object decreases as we move away from its center. It follows an inverse square law, which means that the gravitational acceleration is inversely proportional to the square of the distance. Mathematically, it can be expressed as:
g = G * M / r^2
Where:
- g is the gravitational acceleration
- G is the gravitational constant
- M is the mass of the Earth
- r is the distance from the center of the Earth
In this case, we are given that the gravitational acceleration at the desired point is 1/45 of its value at the Earth's surface. This means that the acceleration (g) at the desired point is equal to 1/45 times the acceleration at the Earth's surface (g0).
1/45 * g0 = G * M / r^2
We can rearrange the equation to solve for the distance (r):
r^2 = (G * M) / (1/45 * g0)
r^2 = 45 * (G * M) / g0
r = sqrt(45 * (G * M) / g0)
To calculate the distance, we need to know the values of the gravitational constant (G), the mass of the Earth (M), and the acceleration at the Earth's surface (g0). The gravitational constant, G, is approximately 6.67430 x 10^-11 m^3 kg^-1 s^-2. The mass of the Earth, M, is approximately 5.972 × 10^24 kg. The acceleration at the Earth's surface, g0, is approximately 9.8 m/s^2.
Plugging in these values into the equation, we can find the distance from the Earth's center to the desired point where the gravitational acceleration is 1/45 of its value at the Earth's surface.