Question

You are explaining why astronauts feel weightless while orbiting in the space shuttle. Your friends respond that they thought gravity was just a lot weaker up there. Convince them and yourself that it isn't so by calculating the acceleration of gravity 248 km above the Earth's surface in terms of g. (The mass of the Earth is 5.97 1024 kg, and the radius of the Earth is 6380 km.)

Answer 1

The acceleration of gravity is inversely proportional to the square of the distance from the center of the Earth.

If the radius of the Earth is 6,380 km and the shuttle used to orbit 248 km above the surface, then

-- the distance from the center when you're on the surface is  6,380 km

-- the distance from the center when you're in orbit on the shuttle is  6,628 km

-- the acceleration of gravity at that altitude is

                  (6,380 / 6,628)²  times 'g' at the surface

             =      (0.963)²    times 'g' at the surface

             =        92.7%  of 'g' at the surface

             =  about  9.1 m/s²  .

(Figuring it out this way, we don't need to know the Earth's mass.)

Answer 2

Answer:

The acceleration of gravity at 248 km above the Earth's surface is $9.08 m/s^2$.

Explanation:

Mass of an object at 248 km above earth = m

Mass of earth = 5.97 1024 kg

Radius of the earth = 6380 km

Distance between earth and object,d= r + 248 km = 6628 km

Gravitational constant = G = $6.674\times 10^{-11} m^3/ kg s^2$

Gravitational force between object and earth:

$F'=G\frac{Mm}{d^2}$

Weight of the object at 248 km above earth : W

W' = mg'

W' = F'

$mg'=G\frac{Mm}{d^2}$

$g'=G\frac{M}{d^2}$.....[1]

Weight of the object on the surface of the earth:

W = mg

Gravitational force between on the surface earth:

$F=G\frac{Mm}{r^2}$

F = W

$g=G\frac{M}{r^2}$....[2]

Dividing [1] by [2]

$\frac{g'}{g}=\frac{G\frac{M}{d^2}}{G\frac{M}{r^2}}=\frac{r^2}{d^2}$

$\frac{g'}{g}=\frac{(6,380 km)^2}{(6,628 km)^2}=0.9267$

$g'=0.9267g = 0.9267\times 9.8 m/s^2=9.08 m/s^2$

The acceleration of gravity at 248 km above the Earth's surface is $9.08 m/s^2$.