Question

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.8×10⁴ m/s when at a distance of 2.6×10¹¹ m from the center of the sun, what is its speed when at a distance of 5.2×10¹⁰ m

v=…..m/s

Answer 1

Magnitude of torque is given by:rF sin where r and F are magnitudes of position and force vectors respectively, is the angle between position and force vectors.

Also, magnitude of angular momentum is given by: mvr sin where m is mass, r and v are magnitudes of position and velocity vectors respectively, is the angle between position and velocity vectors.

For the motion of comet around the sun, we see that position vector is radially outward while gravitational force,being attractive, is radially inward. So,angle between force and position vectors is 180 degrees.So,magnitude of torque on comet about the sun=rFsin180=0.So,there is no torque about the sun on the comet=> angular momentum of the comet is conserved about the sun.

Also, velocity vector is tangential while position vector is radial,so angle between position and velocity vectors is 90 degrees.So,angular momentum=mvr sin90=mvr,where m is mass of comet , v is magnitude of its velocity, r is the distance between the sun and comet.

Initially,v=1.8*10⁴ m/s and r=2.6*10¹¹ m. So, initial angular momentum=m*1.8*10⁴*2.6*10¹¹ kg-m²/s where m is mass of comet in kg.

Finally,r=5.2*10¹⁰ m=>Final angular momentum=m*v*5.2*10¹⁰,where m is mass of comet in kg, v is its velocity at this distance in m/s

So,initial angular momentum=final angular momentum=>m*1.8*10⁴*2.6*10¹¹=m*v*5.2*10¹⁰

=>v=1.8*10⁴*2.6*10¹¹/(5.2*10¹⁰)=9*10⁴ m/s

So,required velocity=9*10⁴ m/s