Explain why Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km

As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite ? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0Xx10^24 kg , radius = 6400 km.

The top of the atmosphere is at about 400 KV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm^-1 . Why then do we not get an electric shock as we step out of hte house into the open? Assume the house to be steel cage so there is no field inside?

Answer the following: The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 Vm^-1 . Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!)

The density of air in atmoshpere decreases with height and can be expressed by the relaton rho = rho_0e^(-alphah where rho_0 is the density at sea level , alpha is a constant and h is the height, Calculate the atmospheric pressure at sea level. Assume g to be constatn. The numerical values of constants are : g = 9.8 ms^-2 , rho_0 = 1.3 kg m^-3 , alpha = 1.2 xx 10^-4 m^-1

If the value of acceleration due to gravity at a height of 100 km from the surface of earth is g ',then at what depth ,value of acceleration due to gravity wioll again be g'?