How far from the center of the earth must a satellite be placed so that the acceleration due to

gravity is half the value at the surface? (Hint: the diameter of the earth is 1.3x107 m.)

the acceleration at 1 radius is 9.8

Since F=k/r^2, half the value means 1/√(1/2) = √2 times the radius.

To find the distance from the center of the Earth at which the acceleration due to gravity is half the value at the surface, we can start by using Newton's law of universal gravitation:

F = G * (m1 * m2) / r^2

Where:
F is the force of gravity between two objects,
G is the gravitational constant (approximately 6.67430 × 10^-11 N m^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 satellite is the "m2" object and the Earth is the "m1" object.

The acceleration due to gravity is given by the equation:

g = G * m1 / r^2

Where:
g is the acceleration due to gravity on the surface of the Earth (approximately 9.8 m/s^2).

To find the distance from the center of the Earth at which the acceleration due to gravity is half the value at the surface (g/2), we can rearrange the equation:

g / 2 = G * m1 / r^2

Rearranging further, we have:

r^2 = G * m1 / (g / 2)

Simplifying, we get:

r = sqrt((2 * G * m1) / g)

Now we can substitute the values:

The Earth's diameter is given as 1.3x10^7 m, which means the radius is half of that: 6.5x10^6 m.

The mass of the Earth (m1) is approximately 5.97x10^24 kg.

Plugging in these values into the equation, we get:

r = sqrt((2 * 6.67430 × 10^-11 N m^2 / kg^2 * 5.97x10^24 kg) / (9.8 m/s^2 / 2))

After evaluating the equation, we find:

r ≈ 1.20x10^7 m

Therefore, a satellite must be placed approximately 1.20x10^7 meters from the center of the Earth to experience half the acceleration due to gravity at the surface.