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
[tex]F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }[/tex]
Where [tex]F_{G}[/tex] = Gravitational force
G = Gravitational constant = 6.67×10⁻¹¹[tex]\frac{Nm^{2} }{kg^{2} }[/tex]
m₁ = mEarth = mass of Earth = 6×10²⁴ kg
m₂ = The other mass which is acted upon by [tex]F_{G}[/tex] 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 [tex]F_{G}[/tex] = [tex]\frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} }[/tex] = 9.231 N
Therefore the acceleration due to gravity = [tex]F_{G}[/tex] /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
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
