A planet of mass 5 Times 10^24 kg is at location (5 Times 10^11, -5 Times 10^11,

Question

A planet of mass 5 Times 10^24 kg is at location (5 Times 10^11, -5 Times 10^11, 0) m. A star of mass 3 Times 10^30 kg is at location (-4 Times 10^11, 4 Times 10^11, 0) m. It will be useful to draw a diagram of the situation, including the relevant vectors. What is the relative position vector r pointing from the planet to the star? What is the distance between the planet and the star? What is the unit vector r in the direction of r? What is the magnitude of the force exerted on the planet by the star? What is the magnitude of the force exerted on the star by the planet? What is the force (vector) exerted on the planet by the star? (Note the change in units.) What is the force (vector) exerted on the star by the planet? (Note the change in units.)

Explanation / Answer

P = <5e11, -5e11,0 > m
S = <-4e11, 4e11,0> m

a. S - P = <9e11, -9e11,0> m

b. By Pythagoras, the length of a vector <a,b,c> is sqrt(a^2+b^2+c^2). So for S - P, we get

| S - P | = sqrt(9e11^2 + (-9e11)^2 + 0^2) = sqrt(81e22 + 81e22 + 0) = sqrt(162e22) = 12.73e11

c. The unit vector is just the relative vector, shortened by its length, so that the resulting length is 1 (therefore "unit"). So
(S - P) / |S - P| = <9e11, -9e11,0> m / 12.73e11 = <0.71, -0.71, 0> m

d. F=(G M1 M2)/r^2 = (6.67*10^-11*5*10^24*3*10^30)/(12.73e11)^2 = 6.17*10^20 N

e. Force on both planet and star will be equal = 6.17*10^20 N

f. F(on planet) =  6.17*10^20 N

g. F(on star) =  6.17*10^20 N

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.