Respuesta :
Answer:
The classification of that same issue in question is characterized below.
Explanation:
The given values are:
Speed, V = 5800 m/s
(a)...
As we know,
⇒ [tex]R=\frac{Gm}{V^2}[/tex]
On putting the estimated values, we get
⇒ [tex]=\frac{6.673\times 10^{-11}\times 10^{24} }{5800^2}[/tex]
⇒ [tex]=1.185\times 10^{-6}-11+24[/tex]
⇒ [tex]=1.185\times 10^7 \ m[/tex]
Now,
Time, [tex]T=2\pi\times \sqrt{\frac{r^3}{Gm} }[/tex]
⇒ [tex]=2\pi \times (\frac{1.185\times 10^{7^{3}}}{6.673\times 10^{-11}\times 5.9742\times 10^{24} } )[/tex]
⇒ [tex]=2\pi \sqrt{0.0417\times 10^8}[/tex]
⇒ [tex]= 12824.12 \ sec[/tex]
So that Time will be "3.56 hours".
(b)...
⇒ [tex]F=ma[/tex]
and,
⇒ [tex]F=\frac{GMm}{r^2}[/tex]
So,
⇒ [tex]ma=\frac{GMm}{r^2}[/tex]
Now,
⇒ [tex]a=\frac{Gm}{r^2}[/tex]
On pitting the estimated values, we get
⇒ [tex]=\frac{6.673\times 10^{-11}\times 5.9742\times 10^{24}}{(1.185\times 10^7)^2}[/tex]
⇒ [tex]=28.38\times 10^{-1}[/tex]
⇒ [tex]=2.838 \ m/s^2[/tex]
(a) The time eriod of revolution is 3.54h
(b) The radial acceleration of the satellite is 2.83 m/s²
Orbit of a satellite:
Let the mass of the satellite be m, amd M be the mass of the earth.
Orbital speed of the satellite is v = 5800 m/s.
(a) In the orbit, the gravitational force is balanced by the centrifugal force as shown below:
[tex]\frac{mv^2}{R}=\frac{GMm}{R^2}[/tex]
where R is the radius of the orbit
[tex]R =\frac{GM}{v^2}\\\\R=\frac{6.67\times10^{-11}\times5.97\times10^{24}}{5800^2}[/tex]
R = 1.18 × 10⁷ m
Now, the time period of a satelite in orbit is given by:
[tex]T=2\pi\sqrt{\frac{R^3}{GM} }\\\\T=2\times3.14\times\sqrt{\frac{(1.18\times10^7)^3}{6.67\times10^{-11}\times5.97\times10^{24}}}[/tex]
T = 12744 s
T = 3.54 h
(b) The radial acceleration is given by:
ma = GMm/R²
a = GM/R²
a = (6.67×10⁻¹¹)(5.97×10²⁴) / (1.18×10⁷)
a = 2.83 m/s²
Learn more about gravitational force:
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