Question

A planet orbits a star in an elliptical orbit. At a particular instant the momentum of the planet is <-3.9 x 10^29, -1.6 x 10^29, 0> kg*m/s, and the force on the planet by the star is <-3.3 x 10^22, -1.9 x 10^23, 0> N. Find FlI and FI_ (Express your answers in vector form.)

Answer 1

The component of the parallel force is $\F_x = (-9.53e22, -3.91e22)kg.m/s$ and the component of the perpendicular force is $F_y = 6.23e22, -1.51e23)N$

To solve this problem, we need to find the magnitude of the momentum of the particle.

Magnitude of Momentum of Particle

This can be calculated as

$|F| = \sqrt{(-3.9*10^2^9)^2 + (-1.6*10^2^9)^2}\\|F| = 4.21*10^2^9m/s$

The unit vector of the momentum is

$p = \frac{p}{|P|} \\ p = \frac{-3.9*10^2^9, -1.6*10^2^9}{4.2*10^2^9}\\ p = (-0.928, -0.38)kg.m/s$

The product of the magnitude of the momentum and the unit vector is

$F.P = [(-3.3*10^2^2, -1.9*10^2^3)N] * [(-0.928, -0.38)]kg.m/s\\F.P = (3.0624*10^2^2 + 0.722*10^2^3)\\F.P = 1.028*10^2^3$

• Let the parallel component of the forces be represented by x
• Let the perpendicular component of the forces be represented by y

Parallel Component of the Forces

The parallel component can be calculated as

$F_x = (F.P)P\\F_x = [(1.028e23)(-0.928, -0.38) kg.m/s}\\\\F_x = (-9.53e22, -3.91e22)kg.m/s$

Perpendicular Component of the Forces

The perpendicular component of the forces is

$F_y = F - F_x\\F_y = (-3.3*10^2^2, -1.9*10^2^3)N - (-9.53e22 - 3.91e22)\\F_y = 6.23e22, -1.51e23)N$

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