Question

Find the radius R of the orbit of a geosynchronous satellite that circles the earth. (Note that R is measured from the center of the earth, not the surface.) You may use the following constants: The universal gravitational constant G is 6.67×10−11N⋅m2/kg2. The mass of the earth is 5.98×1024kg. The radius of the earth is 6.38×106m.

Answer 1

Answer:

R^3 = GM / ω^2

R^3 = (6.67 * 10^-11) * (5.98 * 10^24) / (0.00007272)^2

R^3 = 7.54 * 10^22

R = 42,251,269

R = 4.225 * 10^7 m

4.225 * 10^7

Explanation:

Answer 2

Answer:

The radius R of the orbit of the geosynchronous satellite is $4.225x10^{7} m$

Explanation:

By means of the equation of the orbital speed, the orbital period can be known:

$v = \frac{2 \pi r}{T}$  (1)

Where r is the orbital radius and T is the orbital radius

r can be isolated from equation 1:

$T \cdot v= 2 \pi r$  

$r = \frac{T \cdot v}{2 \pi}$ (2)

Notice that, from equation 2 is necessary to find the velocity before the orbital radius can be determined. That can be done through the Law of Universal gravity.

$F = G\frac{M \cdot m}{r^{2}}$ (3)

Then, replacing Newton's second law in equation 3 it is gotten:

$m \cdot a= G\frac{M \cdot m}{r^{2}}$ (4)

However, a is the centripetal acceleration since it is a circular motion:

$a = \frac{v^{2}}{r}$  (5)

Replacing equation 5 in equation 4 it is gotten:

$m \frac{v^{2}}{r} = G\frac{M \cdot m}{r^{2}}$

$m \cdot v^{2} = G \frac{M \cdot m}{r^{2}}r$

$m \cdot v^{2} = G \frac{M \cdot m}{r}$

$v^{2} = G \frac{M \cdot m}{rm}$

$v^{2} = G \frac{M}{r}$

$v = \sqrt{\frac{G M}{r}}$  (6)

Where v is the orbital speed, G is the gravitational constant, M is the Earth mass, and r is the Earth radius.

Finally, equation 6 can be replaced in equation 2:

$r = \frac{T \cdot v}{2 \pi}$

$r = \frac{T \sqrt{\frac{G M}{r}}}{2 \pi}$

$r^{2} = \frac{T^{2} G M}{4 \pi^{2} r}$

$r^{3} = \frac{T^{2} G M}{4 \pi^{2}}$

$r = \sqrt[3]{\frac{T^{2} G M}{4 \pi^{2}}}$ (7)

Where r is the orbital radius, T is the orbital period, G is gravitational constant and M is the Earth mass.

Since it is a geosynchronous satellite, it will have the same orbital period of the Earth (24 hours).

It is necessary to express the period in seconds:

$T = 24 h x \frac{3600 s}{1 h}$ ⇒ $86400 s$

$T = 86400 s$

Hence, all the values can be replaced in equation 7:

$r = \sqrt[3]{\frac{(86400 s)^{2}(6.67x10^{-11}N.m^{2}/kg^{2})(5.98x10^{24}kg)}{4 \pi^{2}}}$

$r = \sqrt[3]{7.54x10^{22}} m^{3}$

$r = 4.225x10^{7} m$

So the radius R of the orbit of the geosynchronous satellite is $4.225x10^{7} m$