Question

At an orbital height of 400 k m find the orbital period of the satellite

Answer 1

Answer: 4 sec

Explanation:

T= 2pi under root (R+h) ^3 /GM

T= 6.28 under root (6400+400)^3 /6.67 * 10^-11

T=6.28 under root 314432*10^6 /6.67*10^-11

T= 6.28 under root 47141229385.3*10^-11

T= 6.28 under root 0.4

T =6.28 *0.6324

T=3.97 ~ 4 sec

Answer 2

Given that,

Height at which the satellite is placed, h = 400 km

To find,

The orbital period off the satellite

Solution,

The orbital time period of the satellite is given by the following relation as follows :

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

Here,

G is universal gravitational constant

M is mass of Earth

r is distance of satellite form earth, r = R + h, R is radius of Earth

$T=2\pi \sqrt{\dfrac{r^3}{GM}}\\\\T=2\pi\sqrt{\dfrac{\left(6.8\cdot10^{6}\right)^{3}}{6.67\cdot10^{-11}\cdot6\cdot10^{24}}}\\\\T=5569.35\ s$

or

T = 1.54 hour

So, the orbital period of the satellite is 1.54 hours.

Learn more,

At an orbital height of 400 k m find the orbital period of the satellite  ?

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