Protons in cosmic-rays strike the earth's over the earth's surface of `10` protons `m^(-2)s^(-1)`. What dioes the total current the earth receive from beyond its atmosphere in the form of incident consmic ray proton ? The earth's radius is `6.4 xx 10^(6)m`

Find the mass of earth (Given the radius of earth R= 6.4 xx 10^6 m) .

In cosmic rays 0.15 protons cm^(-2) s^(-1) are entering the Earth's atmosphere . If the radius of the Earth is 6400 km, the current received by the Earth in the form of cosmic rays is nearly.

Find the escape velocity of a body from the surface of the earth. Given radius of earth = 6.38 xx 10^(6) m .

At what height from the surface of earth will the value of g becomes 40% from the value at the surface of earth. Take radius of the earth = 6.4 xx 10^(6) m .

How far away from the earth does the acceleration due to gravity become 10% of its value on earth's surface? Radius of earth = 6.37 xx 10^(6) m .

How far away from the earth does the acceleration due to gravity become 10% of its value on earth's surface? Radius of earth = 6.37 xx 10^(6) m .

The surface density of charge on the earth's surface is 2.65xx10^(-9)C//m^(2) . If the radius of the earth is 6400 km then the charge carried by the earth is

A proton is to circulate the earth along the equator with a speed of 1.0 xx 10^(7) m/s. The minimum magnetic field which should be created at the equator for this purpose. (The mass of proton =1.7 xx 10^(-27) kg and radius of earth = 6.37 xx 10^(6) m.) will be (in "Wb/m"^(2) )

How much below the surface of Earth does the acceleration due to gravity become 70% of its value on the surface of Earth. Radius of Earth = 6.4 xx 10^(6) m .