Question

4. Suppose a newborn panda bear of mass 98.0 kg is exhibiting uniform circular motion while

orbiting a neutron star of mass 1.50*10^21 kg and radius 3.50*10^5 m. The distance that the

bear maintains from the surface of the neutron star is 50:0 km. What is the magnitude of the

baby panda's tangential velocity?

Answer 1

Given,

mass of the new born panda bear, m = 98 kg

mass of the neutron star, M = 1.5 * 10²¹ kg

radius of the star, r = 3.50 * 10⁵ m = 350 * 10³ m

distance of panda bear from the surface of the star, d = 50 km = 50 * 10³ m

Hence distance of panda bear from the center of the neutron star, R = r + d

                                                                                                               = 350 * 10³ + 50 * 10³

                                                                                                               = 400 * 10³ m

Now,

Gravitational force between the neutron star and panda bear, F₁ = GMm / R²

Let the tangential velocity of panda bear be v

Due to the circular motion, centrifugal force, F₂ = mv² / R

for circular motion, F₁ = F₂

=> GMm / R² = mv² / R

=> v² = GM / R

=> v =

put the values,

=> v =

         = = 4.4746 * 10² m/s

=> v = 447.46 m/s

thus tangential velocity of the panda bear is 447.46 m/s