Question

At what altitude should a satellite be placed into circular orbit so that its orbital period is 48.0 hours? The mass of the earth is 5.976 × 1024 kg and the radius of the earth is 6.378 × 106 m.

Answer 1

Answer:

$60.689\times 10^6\ m$

Explanation:

Given:

Mass of Earth (M) = $5.976\times 10^{24}\ kg$

Radius of Earth (R) = $6.378\times 10^{6}\ m$

Time period of the satellite (T) = 48.0 hours

Converting time period from hours to seconds using the conversion factor, we get:

1 hour = 3600 s

So, 48.0 hours = 48.0 × 3600 = 172800 s

Let the distance of the satellite's orbit from Earth's center be 'r'.

We know that, the time period of the circular orbit is given by the formula:

$T=2\pi\sqrt{\frac{r^3}{GM}}$

Where, 'G' is the universal gravitational constant = $6.674\times 10^{-11}\ m^3kg^{-1}s^{-2}$

Rewriting in terms of 'r', we get:

$\frac{T}{2\pi}=\sqrt{\frac{r^3}{GM}}$

Squaring both sides, we get:

$(\dfrac{T}{2\pi})^2=\dfrac{r^3}{GM}\\\\\\r^3=\dfrac{GMT^2}{4\pi^2}\\\\\\r=\sqrt[3]{\dfrac{GMT^2}{4\pi^2}}$

Plug in the given values and solve for 'r'. This gives,

$r=\sqrt[3]{\dfrac{6.674\times 10^{-11}\times 5.976\times 10^{24}\times (172800)^2}{4\pi^2}} \\\\\\r=67.067\times 10^{6}\ m$

Now, altitude is measured from the surface of Earth. So, the altitude of the satellite's orbit is given by subtracting the radius of Earth from the total radial distance.

So, altitude (h) = Total radial distance (r) - Earth's radius (R)

$h=(67.067-6.378)\times 10^{6}\ m\\\\h=60.689\times 10^{6}\ m$

Therefore, the satellite's orbit must be placed at an altitude of $60.689\times 10^6\ m$