Question

Determine the acceleration due to gravity for low Earth orbit (LEO) given: MEarth = 6.00 x 1024 kg, rEarth = 6.40 x 106 m, G = 6.67 x 10-11 m3kg-1s-2, and LEO is 400 km above Earth's surface.

Answer 1

Answer:9.8m/s^2

Explanation:

Acceleration due to gravity(g)=?

Mass of earth(m)=6x10^24kg

G=6.67x10^(-11)

Radius of earth(r)=6.4x10^6m

g=(mxG)/r^2

g=(6x10^24x6.67x10^(-11))/(6.4x10^6)

g=(4.002x10^14)/(4.096x10^13)

g=9.8m/s^2

Answer 2

Answer:

The answer to the question is as follows

The  acceleration due to gravity for low for orbit is  9.231 m/s²

Explanation:

The gravitational force is given as

$F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }$

Where $F_{G}$ = Gravitational force

G = Gravitational constant = 6.67×10⁻¹¹$\frac{Nm^{2} }{kg^{2} }$

m₁ = mEarth = mass of Earth = 6×10²⁴ kg

m₂ = The other mass which is acted upon by  $F_{G}$ and = 1 kg

rEarth = The distance between the two masses = 6.40 x 10⁶ m

therefore at a height of 400 km above the erth we have

r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m

and  $F_{G}$ = $\frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} }$ = 9.231 N

Therefore the acceleration due to gravity =  $F_{G}$ /mass  

9.231/1 or 9.231 m/s²

Therefore the acceleration due to gravity at 400 kn above the Earth's surface is  9.231 m/s²