Answer:
Let the receiver be at point A and source be at B, Re-transmission from a satellite Given, velocity of waves \[3\times {{10}^{8}}m/s\] Time to reach a receiver \[=4.04\text{ }ms=4.04\times {{10}^{-3}}s\] Let the height of satellite is, \[{{h}_{s}}=600km\] Radius of Earth = 6400 km Size of transmitting antenna = hf We know that, \[\frac{\text{Distance travelled by wave}}{\text{Time}}\text{=Velocity}\,\,\text{of}\,\,\text{waves}\] \[\Rightarrow \] \[\frac{2x}{4.04\times {{10}^{-3}}}=3\times {{10}^{8}}\] or \[x=\frac{3\times {{10}^{8}}\times 4.04\times {{10}^{-3}}}{2}\] \[=6.06\times {{10}^{5}}=606km\] Using Pythagoras theorem, \[{{d}^{2}}={{x}^{2}}-h_{s}^{2}={{(606)}^{2}}-{{(600)}^{2}}=7236\] or d = 85.06 km So, the distance between source and receiver \[=2d=2\times 85.06=170\,km~\] The maximum distance covered on ground from the transmitter by emitted EM waves, \[d=\sqrt{2R{{h}_{T}}}\]or \[{{d}^{2}}/2R={{h}_{T}}\] or size of antenna, \[{{h}_{T}}=\frac{7236}{2\times 6400}=0.565\,km=565m\]
You need to login to perform this action.
You will be redirected in
3 sec