The gravitational attraction between two objects with masses m 1 and m 2, separa

Question

The gravitational attraction between two objects with masses m1 and m2, separated by distance x, is F=Gm1m2/x2, where G is the gravitational constant.

1) How much work is done by gravity when the separation changes from x1 to x2? Assume x2<x1.

2) If one mass is much greater than the other, the larger mass stays essentially at rest while the smaller mass moves toward it. Suppose a 1.5 × 10^13 kg comet is passing the orbit of Mars, heading straight for the sun at a speed of 3.7 × 10^4 m/s. What will its speed be when it crosses the orbit of Mercury? The orbit of Mars is 2.28×10^8 km, the orbit of Mercury is 57.9×10^6 km, and G = 6.67 × 10^11 Nm2/kg2.

Explanation / Answer

1) work done by gravity is W = U1 - U2

U1 = initial gravitational potential energy = G*m1*m2/x1

U2 = final gravitational potential energy = G*m1*m2/x2

then

W = U1 - U2

W = (G*m1*m2/x1)-(G*m1*m2/x2)

2) using law of conservation of energy

K1 + U1 = K2 + U2

(0.5*m*u1^2) - (G*m*M/x1) = (0.5*m*v^2)-(G*m*M/x2)

here x2= 57.9*10^9 m

x1 = 2.28*10^11 m

(0.5*1.5*10^13*(3.7*10^4)^2) - ((6.67*10^-11*1.5*10^13*1.99*10^30)/(2.28*10^11)) = (0.5*1.5*10^13*v^2)-((6.67*10^-11*1.5*10^13*1.99*10^30)/(57.9*10^9))

v == 6.92*10^4 m/sec

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