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)

Respuesta :

Answer:

[tex]a=8.92m/s^{2}[/tex]

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:

[tex]F=\frac{GMm}{r^{2}}[/tex]

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

[tex]F=ma[/tex],

here, making a substitution we get

[tex]ma=\frac{GMm}{r^{2}}\\\\a=\frac{GM}{r^{2}}[/tex]

The distance between columbia and the earth's center is

[tex]r=6.4*10^{6}+0.3*10^{6}=6.7*10^{6}m[/tex].

Now, computing the acceleration:

[tex]a=\frac{(6.67*10^{-11})(6*10^{24})}{(6.7*10^{6})^{2}}\\\\a=8.92m/s^{2}[/tex]