At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.790 m/s^2 if the acceleration due to gravity at the surface has magnitude .
To find the distance above the surface of the Earth where the acceleration due to gravity is 0.790 m/s^2, we need to make use of Newton's Law of Universal Gravitation.
Newton's Law of Universal Gravitation states that the gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
The acceleration due to gravity at the surface of the Earth is approximately 9.8 m/s^2. Let's call this value g.
We can use the following equation to calculate the distance above the surface of the Earth where the acceleration due to gravity is 0.790 m/s^2:
g' = (G * M) / (r + h)^2
Where:
g' is the acceleration due to gravity at a distance h above the Earth's surface
G is the gravitational constant (approximately 6.674 * 10^-11 N m^2 / kg^2)
M is the mass of the Earth
r is the radius of the Earth
h is the distance above the surface of the Earth
Since we know that g is 9.8 m/s^2 and g' is 0.790 m/s^2, we can set up the following equation:
0.790 = (6.674 * 10^-11 * M) / (r + h)^2
From here, we need to rearrange the equation to solve for h:
h = √[(6.674 * 10^-11 * M) / (0.790 * (r + h)^2)] - r
Finding the exact value of h requires knowing the values for M and r, which are the mass and radius of the Earth respectively. However, we can calculate an approximation if we assume that M is approximately 5.972 × 10^24 kg and r is approximately 6,371 km.
Using these rough estimations, you can substitute the known values into the equation and solve for h:
h = √[(6.674 * 10^-11 * 5.972 × 10^24) / (0.790 * (6,371,000 + h)^2)] - 6,371,000
Please note that this is an approximation, and for an accurate value of h, the precise values of M and r should be used.