A planet of mass 4 10 24 kg is at location < 5 x 10 11 , 4 x 10 11 , 0> A star o
ID: 1406742 • Letter: A
Question
A planet of mass 4 1024 kg is at location <5 x 1011, 4 x 1011, 0> A star of mass 3 1030 kg is at location m.<6 x 1011, 4 x 1011, 0> It will be useful to draw a diagram of the situation, including the relevant vectors.NOTE: I placed the arrow beside the letter to show it was a vector.
(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 hat) 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? (Note the change in units.)
F->on planet =
x 1020 N
(g) What is the force (vector) exerted on the star by the planet? (Note the change in units.)
F->on star =
x 1020 N
Answers to a similar problem with different values are, a)-9.0e+11(i hat)+9.00 e + 11(j hat)
i just dont know how to do find them myself. b. 1.27e + 12
c. -7.07i(hat) + 0.707j(hat)
d. 4.94e+20
e. 4.94e+20
f. -3.49i(hat) + 3.49j(hat)
g. 3.49i (hat) -3.49j(hat)
Explanation / Answer
planet <5 x 10^11, 4 x 10^11, 0>
star <6 x 10^11, 4 x 10^11, 0>
(a) relative position vector = <6 x 10^11, 4 x 10^11, 0> - <5 x 10^11, 4 x 10^11, 0>
= < - 11 x 10^11, 8 x 10^11, 0 >
(b) distance between planet and star = sqrt( (11 x 10^11)^2 + (8 x 10^11)^2) = 1.36 x 10^12 m
(c) unit vector = < - 11 x 10^11, 8 x 10^11, 0 > / 1.36 x 10^12 = < - 0.8088, 0.588, 0 >
(d) Force = G M1 M1 / d^2 = 6.674 x 10^-11 x 10^24 x 10^30 / (1.136 x 10^12)^2 = 5.17 x 10^19 N
(e) Force = G M1 M1 / d^2 = 6.674 x 10^-11 x 10^24 x 10^30 / (1.136 x 10^12)^2 = 5.17 x 10^19 N
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