Calculate the velocity with which a body must be thrown vertically upward from the surface of the earth so that it may reach a height of `10R`, where `R` is the radius of the earth and is equal to `6.4 xx 10^(6)m.` (Given: Mass of the earth `= 6 xx 10^(24) kg`, gravitational constant `G = 6.7 xx 10^(-11) N m^(2) kg^(-2)`)
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The gravitational potential of a body of mass `m` on earth's surface is `U(R)=-(GMm)/R` where `M` is the mass of the earth (supposed to be concentrated at its centre) and `R` is the radius of the earth (distance of the particle from the centre of the earth). The gravitational energy of the same body at a height `10R` from earth's surface, i.e, at a distance `11R` from earth's centre is `U(11R)=-(GMm)/R` Therefore, change in potential energy, `U(11R)-U(R)=-(GMm)/(11R)-(-(GMm)/R)=10/11(GMm)/R` this difference must come from the initial kinetic energy given to the body in sending it to that height. Now, suppose the body is throuwn up with a vertical speed `v`, so that its initial kinetic energy is `1/2mv^(2)`. Then `1/2mv^(2)=10/11(GMm)/R` or `v=sqrt(20/11 (Gmm)/R)` putting the given values: `v=sqrt((20xx(6.7x10^(-11)Nm^(2)kg^(-2))xx(6xx10^(24)kg))/(11(6.4xx10^(6)m))` `=1.07xx10^(4)ms^(-1)`
Radius of the earth is 6.40 xx 10^6 m. Find the diameter of the earth.
A rocket is fired vertically with a speed of 5 km s^(-1) from the earth's surface. How far from the earth does the rocket go before returning to the erath? Mass of the earth = 6.0 xx 10^(24) kg , mean radius of the earth = 6.4 xx 10^(6) m. G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) .
What is the work done in taking an object of mass 1kg from the surface of the earth to a height equal to the radius of the earth ? ( G = 6.67 xx 10^(-11) Nm^(2)//Kg^(2) , Radius of the earth = 6400 km, Mass of the earth = 6 xx 10^(24) kg .)
Calculate the height above the earth at which the geostationary satellite is orbiting the earth. Radius of earth = 6400km. Mass of earth = 6 xx 10^(24) kg. G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) .
Calculat the binding energy of the - sum system . Mass of the earth =6xx 10 ^(24) kg , mass of the sun = 2 xx 10^(30) kg, distance between the earth and the sun = 1.5 xx 10^(11) and gravitational constant = 6.6 xx 10 ^(-11) N m^(2) kg^(2)
What is the value of g at a height 8848 m above sea level. Given g on the surface of the earth is 9.8 ms^(-2) . Mean radius of the earth = 6.37 xx 10^(6) m .
An artificial satellite is revolving in a circular orbit at a height of 1200 km above the surface of the earth. If the radius of the earth is 6400 km, mass is 6 xx 10^(24) kg find the orbital velocity (G = 6.67 xx 10^(-11)Nm^(2)//kg^(2))
A rocket is fired vertically with a speed of 5 kms^(-1) from the earth's surface. How far from the earth does the rocket go before returning to the earth ? Mass of earth= 6.0xx10^(24) kg, mean radius of the earth = 6.4xx10^(6) m, G= 6.67xx10^(-11)Nm^(2)Kg^(-2).
If the acceleration due to gravity on earth is 9.81 m//s^(2) and the radius of the earth is 6370 km find the mass of the earth ? (G = 6.67 xx 10^(-11) Nm^(2)//kg^(2))
Find the gravitational potential due to a body o fmass 10kg at a distance (i) 10m and (ii) 20 m from the body. Gravitational constant G = 6.67 xx 10^(-11) Nm^(2) kg^(-1) .
