A planet of mass 5 multiple 10^24kg is at location (6 multiple 10^11,-5 multiple

Question

A planet of mass 5 multiple 10^24kg is at location (6 multiple 10^11,-5 multiple 10^11,0) m. A star of mass 3 multiple 10^30 kg is at location (-4 multiple 10^11, 5 multiple 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 vector pointing from the planet tot the star? What is the distance between the planet and the star? What is the unit vector r^ in the direction of r vector? What is the magnitude of the force exerted on the planet by the star? |F vector on planet| = N what is the magnitude of the force exerted on the star by the planet? |F vector on star| = N What is the force (vector) exerted on the planet by the star? (Note the change in units.) F vector planet = x10^20 N what is the force (vector) exerted on the star by the planet? (Note the change in units.) F vector star= x 10^20 N

Explanation / Answer

a) r from planet to the star = star coordinate - planet coordiinate

r = (-4e11, 5e11, 0) - (6e11, -5e11, 0)

= -10e11, 10e11, 0

b) magnitude of r |r| = sqrt(10^2 + 10^2 +0) x 10^11

= 14.14 x 10^11 m

c) unit vector = r vector / |r|

= (-0.707, 0.707, 0)

d) F = GMm / r^2 = (6.67 x 10^-11 x 5 x 10^24 x 3 x 10^30) / (14.14 x 10^11)^2

= 5 x 10^20 N

e) it will be same from newton's 3rd law.

F = 5 x 10^20 N

f) Fonplanet = 5 x10^20 (-0.707, 0.707, 0)

= (-3.535, 3.535, 0) x 10^20 N

g) direction will be opposite.

F = (3.535, -3.535, 0) x 10^20 N

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