Question

A star of mass 7 × 10³⁰ kg is located at <6 × 10¹², 3 × 10¹², 0> m. A planet of mass 9 × 10²⁴ kg is located at <5 × 10¹², 5 × 10¹², 0> m and is moving with a velocity of <0.9 × 10⁴, 1.5 × 10⁴, 0> m/s.

a/ During a time interval of 1 × 10⁶ seconds, what is the change in the planet's velocity? (in xyz vector form)

b/ During this time interval of 1 × 10⁶ seconds, what is the change in the planet's position?(in xyz vector form)

Answer 1

The coordinates of star ( 6 x 10^12 , 3 x 10^12 ,0)m

The coordinates of planet ( 5 x 10^12 ,5 x 10^12 ,0)m

So both are in X-Y plane and hence their gravitation attraction will also be in X-Y plane ,since velocity has no z-component, entire motion is constrained to x-y plane.

The magnitude of distance between the two = sqrt [ (1 x 10^12 )^2 + (-2 x 10^12) ^2 ] = sqrt 5 x 10^12 m

The magnitude of force = GMm/r^2

the magnitude of acceleration = GM/r^2 = 6.67 x 10^(-11) nt-m^2/m^2 x 7 x 10^30 Kg / 5 x 10^24 = 9.34 x 10^(-5) m/s^2

= 4.18 x 10^(-5) m/s^2

= -8.32 m/s^2

= 4.18 x 10^(-5) m/s^2 x 1 x 10^6 = 41.8 m/s

= - 8.32 x 10^(-5) m/s^2 x 1 x 10^6 s = -83.2 m/s

Change in vel = ( 41.8 , -83.2,0) m/s

b) V =( 0.9 x 10^4, 1.5 x 10^4) m/s

= 0.9 x 10^4 m/s x 10^6 sec = 90m

= 1.5 x 10^4 m/s x 10^6 sec = 150m