Question

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 1.5×1013 kg comet is passing the orbit of Mars, heading straight for the 1.99×1030 kg sun at a speed of 3.1 ×104 m/s. What will its speed be when it crosses the orbit of Mercury? The orbit of Mars is 2.28×108 km, the orbit of Mercury is 5.79×107 km, and G = 6.67×10-11 N⋅m2/kg2.

Answer 1

Answer:

The speed of orbit when it crosses Mercury, v = 5.4 x 10⁴ m/s

Explanation:

Given,

The mass of the comet, m = 1.5 x 10³³ Kg

The mass of the sun, M = 1.99 x 10³⁰ Kg

The velocity of the comet at the orbit of mars, V = 3.1 x 10⁴ m/s

The radius of orbit of mars, R = 2.28 x 10⁸  km

                                                 = 2.28 x 10¹¹ m

The radius of orbit of mercury, r = 5.79 x 10⁷ Km

                                                      = 5.79 x 10¹⁰ m

The velocity of orbit at a distance R from the sun is given by the formula

                                            $V = \sqrt{\frac{GM}{R} }$

Substituting the given values in the above equation

                                           $V = \sqrt{\frac{6.67X10^{-11}X1.99X10^{30}}{2.28X10^{11}}}$

                                                    = 2.41 x 10⁴ m/s

Given that the velocity of a comet passing the orbit of mars is 3.1 x 10⁴ m/s

The discrepancy in the velocity is 0.7 x 10⁴ m/s

This is due to the high eccentricity of the orbit of a comet.

The velocity of a comet crossing the orbit of mercury is

                                             $v = \sqrt{\frac{6.67X10^{-11}X1.99X10^{30}}{5.79X10^{10}}}$    

                                                     = 4.7 x 10⁴ m/s

Adding the discrepancy to the above value gives

                                                  v =  5.4 x 10⁴ m/s

Hence, the velocity of the comet crossing the orbit of mercury is, v =  5.4 x 10⁴ m/s