Question

If columbia is orbiting at 0.3*10^6 m above the earths surface, what is the accceleratoin of Columbia

due to the Earth’s gravity?
(Radius of Earth = 6.4 x 106 m, mass of Earth = 6.0 x 1024 kg)

Answer 1

`a=8.92m/s^(2)`

Step-by-step explanation:

For this problem, we need to make use of Newton's law of universal gravitation. This law states that two objects attract each other with a force that is inversely proportional to the square of the distance of their centers of mass and directly proportional to the product of their masses. We can write this as:

`F=(GMm)/(r^(2))`

where F is the attractive force, G is the gravitational constant, r is the distance between their centers of mass, and M and m are the masses of the objects.

From here we will let M be the mass of the earth, and m the mass columbia. From Newton's second law, we know that the gravitational force exerted to columbia due to the earth can be written as

`F=ma`,

here, making a substitution we get

`ma=(GMm)/(r^(2))\\\\a=(GM)/(r^(2))`

The distance between columbia and the earth's center is

`r=6.4*10^(6)+0.3*10^(6)=6.7*10^(6)m`.

Now, computing the acceleration:

`a=((6.67*10^(-11))(6*10^(24)))/((6.7*10^(6))^(2))\\\\a=8.92m/s^(2)`