Question

A planet of mass 7 × 10²⁴ kg is at location <2 × 10¹¹, -4 × 10¹¹, 0> m. A star of mass 4 × 10³⁰ kg is at location <-6 ×10¹¹, 6 × 10¹¹, 0> m. It will be useful to draw a diagram of the situation, including the relevant vectors.
(a) What is the relative position vector r→ pointing from the planet to the star?

r→=	< , ,  > m

(b) What is the distance between the planet and the star?

|r→|=

m

(c) What is the unit vector r^ in the direction of r→?

r^=

<, ,  >

(d) What is the magnitude of the force exerted on the planet by the star?

|F→on⁢planet|=

N

(e) What is the magnitude of the force exerted on the star by the planet?

|F→on⁢star|=

N

(f) What is the force (vector) exerted on the planet by the star?

F→on⁢planet=

<, ,  > N

(g) What is the force (vector) exerted on the star by the planet?

F→on⁢star=

<, ,  > N

Answer 1

Given : mass of planet = ms = 7 x 1024 kg

position vector of planet =

mass of star = ms = 4 x 1030 kg

position vector of star =

See the diagram :

P = position of planet

S = position of star

the line of P to origin and S to origin are not collinear.

a) relative position vector pointing from the planet to the star .

therefore, = < -8 x 1011, 10 x 1011, 0>m [Answer]

b) distance between planet and star = ||

[answer]

c) unit vector in the direction of =

[answer]

d) Magnitude of force exerted on the planet by the star = F

. [Answer]

e) Magnitude of the force exerted on the star by the planet is same as Magnitude of force exerted on the planet by the star by Newton's 3rd Law.

therefore, Magnitude of the force exerted on the star by the planet = F = 1.14 x 1021 N [answer]

f) Force in vector form exerted on the planet by the star =

[since, ]

therefore, = <0.712 x 1021, - 0.89 x 1021 , 0> N [Answer]

g) Force in vector form exerted on the star by the planet =

From newton's 3rd law:

= -

Therefore, = <-0.712 x 1021, 0.89 x 1021 , 0> N [Answer]