An artificial satellite circles around the earth at the height where the gravitational force is one fourth of that at earth's surface. If the earth's radius is R the height of the satellite above the surface of the earth is?
r and Re
F = G Mm/r^2 = (.25) G Mm/Re^2
or
.25/Re^2 =1/r^2
.25 r^2 = Re^2
.5 r = Re
r = 2 Re
height = 2Re-Re = Re
To solve this problem, we need to use the concept of gravitational force and the formula for gravitational force between two objects.
The formula for gravitational force between two objects is given by:
F = (G * m1 * m2) / r^2
Where:
F is the gravitational force,
G is the gravitational constant (approximately 6.674 * 10^-11 Nm^2/kg^2),
m1 and m2 are the masses of the two objects, and
r is the distance between the centers of the two objects.
In this case, the two objects are the satellite and the Earth. Let's denote the mass of the satellite as msatellite, the mass of the Earth as mearth, the radius of the Earth as R, and the distance between the satellite and the center of the Earth as h.
At the Earth's surface, the gravitational force is given by:
F1 = (G * mearth * msatellite) / R^2
At the height where the gravitational force is one-fourth, the gravitational force is given by:
F2 = (G * mearth * msatellite) / (R + h)^2
According to the problem, F2 is one-fourth of F1. Therefore, we have the equation:
F2 = (1/4) * F1
Substituting the expressions for F1 and F2, we get:
(G * mearth * msatellite) / (R + h)^2 = (1/4) * (G * mearth * msatellite) / R^2
Canceling out common terms:
1 / (R + h)^2 = 1 / (4 * R^2)
Cross-multiplying:
(R + h)^2 = 4 * R^2
Expanding:
R^2 + 2 * R * h + h^2 = 4 * R^2
Rearranging the terms:
3 * R^2 - 2 * R * h - h^2 = 0
This quadratic equation can be solved to find the value of h.
Using the quadratic formula:
h = [ -(-2 * R) ± sqrt((-2 * R)^2 - 4 * 3 * (-h^2))] / (2 * 3)
Simplifying:
h = [2 * R ± sqrt(4 * R^2 + 12 * h^2)] / 6
h = [R ± sqrt(R^2 + 3 * h^2)] / 3
Solving for h:
3h = R ± sqrt(R^2 + 3h^2)
3h - sqrt(R^2 + 3h^2) = R
Squaring both sides to eliminate the square root:
9h^2 - 6h * sqrt(R^2 + 3h^2) + R^2 = R^2
9h^2 - 6h * sqrt(R^2 + 3h^2) = 0
h(9h - 6 * sqrt(R^2 + 3h^2)) = 0
h = 0 \[ \text{(reject this solution as height cannot be zero)} \] or
9h - 6 * sqrt(R^2 + 3h^2) = 0
Rearranging:
6 * sqrt(R^2 + 3h^2) = 9h
Simplifying:
2 * sqrt(R^2 + 3h^2) = 3h
Squaring both sides:
4 * (R^2 + 3h^2) = 9h^2
4R^2 + 12h^2 = 9h^2
4R^2 - 9h^2 = -12h^2
4R^2 = 3h^2
R^2 = (3/4)h^2
Taking the square root:
R = sqrt(3/4)h
Simplifying:
h = (4/3)R
Therefore, the height of the satellite above the surface of the Earth is (4/3) times the radius of the Earth.
To solve this problem, we need to start with the equation for gravitational force:
F = G * (m1 * m2) / r^2
Where:
F is the gravitational force between two objects,
G is the gravitational constant,
m1 and m2 are the masses of the two objects,
r is the distance between the centers of the two objects.
At the surface of the Earth, the gravitational force (F_surface) acting on an object of mass m is given by:
F_surface = G * (m * M_earth) / R^2
Where:
M_earth is the mass of the Earth.
Considering the satellite at a height above the Earth's surface, the gravitational force acting on the satellite (F_satellite) is given by:
F_satellite = G * (m * M_earth) / (R + h)^2
Where:
h is the height of the satellite above the Earth's surface.
Given that the gravitational force acting on the satellite is one-fourth (1/4) of the gravitational force at the Earth's surface, we can set up the following equation:
F_satellite = (1/4) * F_surface
Substituting the expressions for F_surface and F_satellite into the equation, we have:
G * (m * M_earth) / (R + h)^2 = (1/4) * (G * (m * M_earth) / R^2)
Now, we can cancel out the terms involving G, m, and M_earth from both sides of the equation:
1 / (R + h)^2 = (1/4) / R^2
Next, we can simplify the equation by cross-multiplying:
4 * R^2 = (R + h)^2
Expanding and rearranging the equation, we have:
4 * R^2 = R^2 + 2 * R * h + h^2
Simplifying further:
4 * R^2 - R^2 = h(2 * R + h)
3 * R^2 = h(2 * R + h)
Now, we have a quadratic equation that can be solved for h. Simplifying it further:
3 * R^2 = 2 * R * h + h^2
h^2 + 2 * R * h - 3 * R^2 = 0
This is a quadratic equation in the form of ah^2 + bh + c = 0, where a = 1, b = 2R, and c = -3R^2. We can solve this quadratic equation using the quadratic formula:
h = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we have:
h = (-(2R) ± √((2R)^2 - 4(1)(-3R^2))) / (2(1))
Simplifying further, we get:
h = (-2R ± √(4R^2 + 12R^2)) / 2
h = (-2R ± √(16R^2)) / 2
h = (-2R ± 4R) / 2
The two possible values for h are:
h1 = (2R - 4R) / 2 = -2R / 2 = -R (ignoring the negative solution)
h2 = (2R + 4R) / 2 = 6R / 2 = 3R
Since height cannot be negative, the height of the satellite above the surface of the Earth is 3R.