Question

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.5×10⁴ m/s when at a distance of 2.8×10¹¹ m from the center of the sun, what is its speed when at a distance of 5.3×10¹⁰ m.

Answer 1

Apply conservation of energy to solve this problem.

Means,

Potential Energy (PE) + Kinetic energy (KE) at one distance = PE + KE at the second distance

the PE due to the sun is -GMm/r

where G is the Universal Gravitational Constant = 6.67x10^-11 in MKS units

M = mass of the sun = 2 x 10^30 kg

r = the distance of the comet from the sun

m = mass of the comet, but this appears in each term of the equation, so will cancel out and will not be needed, so we have

-GM/r1 + v1^2/2 = - GM/r2 + v2^2/2 where 1 and 2 refer to the first and second positions;

Plug in the values -

(- 6.67x10^-11 x 2 x 10^30 kg) / (2.8x10^11 m) + (2.5 x 10^4 m/s)^2 / 2 = (-6.67x10^-11 x 2 x 10^30 kg) / (5.3x10^10 m) + v2^2/2

=> -4.76 x 10^8 + 3.125 x 10^8 = -2.517 x 10^9 + v2^2 / 2

=> - 1.635 x 10^8 = - 25.17 x 10^8 + v2^2 / 2

=> v2^2 / 2 = 23.535 x 10^8

=> v2^2 = 2 x 23.535 x 10^8

=> v2 = 6.86 x 10^4 m/s (Answer)