Determine the answer to the equation 30 km/h × 17 h =

Chegg Given that the mean radius of the Moon’s orbit is 3.84 x 108 m and its period is 2.36 x 106 sec, at what altitude above th

alukav5142 [94]

Answer:

The altitude of geostationary satellite is $3.58\times10^{7}\ m$

Explanation:

Given that,

Radius of moon's orbit $r=3.84\times10^{8}\ m$

Time period $T=2.36\times10^{6}\ sec$

We need to calculate the orbital radius of geostationary satellite is

Using formula of time period

$T=\sqrt{\dfrac{4\pi^2}{GM}a^3}$

$a=((\dfrac{GM}{4\pi^2})T^2)^{\dfrac{1}{3}}$

Where, G = gravitational constant

M = Mass of earth

T = time period of geostationary satellite orbit

Put the value in to the formula

$a=((\dfrac{6.67\times10^{-11}\times5.97\times10^{24}}{4\times\pi^2})\times(86160)^2)^{\dfrac{1}{3}}$

$a=4.217\times10^{7}\ m$

We need to calculate the altitude of geostationary satellite

Using formula of altitude

$h = a-R_{e}$

Where, R = radius of earth

a = radius of geostationary satellite

Put the value into the formula

$h =4.217\times10^{7}-6.38\times10^{6}$

$h =35790000\ m$

$h=3.58\times10^{7}\ m$

Hence, The altitude of geostationary satellite is $3.58\times10^{7}\ m$

4 0

4 years ago

distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm.

a_sh-v [17]

Answer:

The speed of the galaxy relative to the Earth is $3.09\times 10^6\ m/s$ .

Explanation:

We have,

(a) Wavelength emitted by light at distant galaxy is 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm. It can be seen that the wavelength of light reduces as it reaches Earth. It is called Red shift. As per Doppler's effect, we can say that the galaxy is receding from the Earth.

(b) Let v is the speed of the galaxy relative to the Earth. It can be given by :

$v=c(\dfrac{\lambda'}{\lambda}-1)\\\\v=3\times 10^8\times (\dfrac{438.6 }{434.1 }-1)\\\\v=3\times 10^8\times (\dfrac{438.6}{434.1}-1)\\\\v=0.0103\cdot3\cdot10^{8}\\\\v=3.09\times 10^6\ m/s$