The force of attraction between the Earth and the Sun is F=(G(Ms)(Me))/r^2, wher

Question

The force of attraction between the Earth and the Sun is F=(G(Ms)(Me))/r^2, where Ms and Me are the masses of the Sun and Earth, G=6.67*10^-11 (N.m^2)/kg^2 is the Universal Gravitation Constant and is the separation of the Earth and Sun. If the Earth suddenly stopped in orbit, it would begin to travel in a straight line toward the center of the Sun. At what speed would the Earth be going when it reached Mercury’s orbital radius?(Hint: Draw a picture with the Sun at the origin and the Earth moving along the x-axis.)

Mass of Sun:1.99*10^30kg

Mass of Earth5.97*10^24kg

Orbital Radius of the Earth1.50*10^11m

Orbital Radius of Mercury 5.79*10^8m

Explanation / Answer

gIVEN

Mass of Sun:1.99*10^30kg

Mass of Earth5.97*10^24kg

Orbital Radius of the Earth1.50*10^11m

Orbital Radius of Mercury 5.79*10^8m

we know that the gravitational force of attraction is  

F = G*M1*m2/r^2

when the earth suddenly stops , moves towards sun in straight line due to conservation of energy here

the work done by the gravitational force of sun on earth = change in kinetic energy of the Earth when it reaches the orbital radius of the Mercury

that is  

G*Ms*Me/(r^2) *r= 0.5*Me(v2^2-v1^2)

here v1 = 0 m/s

v2^2 = 2*G*Ms/r

V2 = sqrt(2*G*Ms/r)

V2 = sqrt((2*6.67*10^-11*1.99*10^30)/(5.79*10^8)) m/s

V2 = 677119.26635 m/s

the speed of the Earth at when it reached the Mercury orbital is 677119.26635 m/s

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