How much work is required to lift a 600-kg satellite to an altitude of 4 * 10^6 m above the surface of the earth? The gravitational force is F=GMm/r^2, where M is the mass of earth, m is the mass of the satellite, r is the distance between them. The radius of the earth is 6.4 * 10^6 m, its mass 6 *10^24 kg. And G = 6.67 *10^-11
To find the work required to lift a satellite to a certain altitude, we can use the formula for gravitational potential energy, which is given by:
G.P.E = mgh
In this case, "m" represents the mass of the satellite, "g" represents the acceleration due to gravity, and "h" represents the vertical height (altitude) above the Earth's surface.
To calculate the gravitational potential energy, we need to find the height "h" above the surface of the Earth. In this case, the satellite is lifted to an altitude of 4 * 10^6 m above the surface of the Earth.
However, we are given the distance "r" between the satellite and the center of the Earth, not the height above the surface. We need to take into account the radius of the Earth to convert this distance into the height above the surface.
The height "h" can be calculated using the formula:
h = r - R
Where "R" is the radius of the Earth, given as 6.4 * 10^6 m.
Plugging in the values:
h = (4 * 10^6 m) - (6.4 * 10^6 m)
= -2.4 * 10^6 m
Since the altitude is above the surface, the height is positive.
Now we can calculate the work done to lift the satellite using the formula:
W = G.P.E = mgh
Given the mass of the satellite is 600 kg, and the value of "g" remains constant at approximately 9.8 m/s^2, the equation becomes:
W = (600 kg) * (9.8 m/s^2) * (2.4 * 10^6 m)
Now, you can calculate the final result with the given values.