Solve Gravitational Field Problems: Earth Satellite

[F.B]

I am stuck on these problems i really don't know how to do them.So can anyone please help me. Can you explain how to do these questions.

A 620 kg satellite above the Earth's surface experiences a gravitational field strength of 4.5 N/kg.

a) Knowing the gravitational field strength at Earth's surface abd Earth's radius, how far above the Earth's surface is the satellite? (Use ratio and proportion.

b) Determine the magnitude of the gravitational force on the satellite.

I know for b) i need a) but i don't seem to get the right answer for a.

for a this is wat i do.

gs=GMs/Rs^2
ge=GMe/Re^2

then i get g s/g e = Ms/Me x (Re/Rs)^2
But i don't get the right answer for it.

 

[whozum]

You don't need (a) for (b).

For (a) you are given the acceleration due to gravity (the local g) and need to find the distance between Earth and the satelite.. just use the equation you have to find the distance..

 

[Fermat]

...
for a this is wat i do.
gs=GMs/Rs^2
ge=GMe/Re^2
then i get g s/g e = Ms/Me x (Re/Rs)^2
But i don't get the right answer for it.

The g-value doesn't depend on the mass of the object experiencing the force. It varies due to the distance between the two masses only.

The gravitational force of attraction is given by,

$$F = \frac{GMm}{r^2}$$

The gravitational field strength, as you put it, is given by,

$$g = \frac{F}{m} = \frac{GM}{r^2}$$

Considering your problem,

$$g_s = \frac{GM_e}{r_s^2}$$
$$g_e = \frac{GM_e}{r_e^2}$$
$$\frac{g_s}{g_e} = \frac{GM_e}{r_s^2} \div \frac{GM_e}{r_e^2}$$
$$\frac{g_s}{g_e} = \frac{r_e^2}{r_s^2}$$

Use the above expression to get the height of the satellite above the Earth's surface, circa 3,000 km