Question

At a particular instant the magnitude of the momentum of a planet is 2.60 × 10^29 kg·m/s, and the force exerted on it by the star it is orbiting is 8.5 × 10^22 N. The angle between the planet's momentum and the gravitational force exerted by the star is 138°. What is the parallel component of the force on the planet by the star?

Answer 1

Answer:

F=(-4.8*10^22,0,0) N

Explanation:

Given  :

We are given the magnitude of the momentum of the planet and let us call this momentum (p_now) and it is given by p_now = 2.60 × 10^29 kg·m/s. Also, we are given the force exerted on the planet F = 8.5 × 10^22 N. and the angle between the planet and the star is Ф = 138°

Solution :

We are asked to find the parallel component of the force F The momentum here is not constant, where the planet moving along a curving path with varying speed where the rate change in momentum and the force may be varying in magnitude and direction. We divide the force here into two parts: a parallel force F to the momentum and a perpendicular force F' to the momentum.  

The parallel force exerted to the momentum will speed or reduce the velocity of the planet and does not change its moving line. Let us apply the direction cosines, we could obtain the parallel force as next  

F=|F|cosФp            (1)

Where the parallel force F is in the opposite direction of p as the angle between them is larger than 90°. Now we can plug our values for 0 and I F I into equation (1) to get the parallel force to the planet  

F=|F|cosФp

 =-4.8*10^22 N*p

As this force is in one direction, we could get its vector as next  

F=(-4.8*10^22,0,0) N

F=(0,-4.8*10^22,0) N

F=(0,0-4.8*10^22) N

The cosine of 138°, the angle between F and p is, is a negative number, so F is opposite to p. The magnitude of the planet's momentum will decrease.