Based on the following data about planet X (which orbits around the Sun): Planet

Question

Based on the following data about planet X (which orbits around the Sun):

Planet X's distance from Sun = 3.6E12 m

Planet X'sradius = 2E6 m

Planet X's mass = 6.8E22 kg

a.) Find gx, the size of the acceleration due to gravity on the surface of Planet X. ______ m/s2

b.) What is the weight of a 10 kg mass on the surface of Planet X? _______N

c.) How long would it take for a ball dropped from a height of 4 m to hit the ground? ______s

d.) At 3 of Planet X's radii above the planet's surface, what is gx? ___________m/s2

e.) For Planet X's orbit around the Sun, please find: its orbital speed: m/s its centripetal acceleration: __________m/s2

f.) How long is a year on Planet X? Express your answers in both seconds and Earth years: ___________s.

_________ Earth years

Explanation / Answer

As given in he question,

Distance from Sun: d = 3.6*10^12 m

Radius of planet: R = 2*10^6 m

Mass of planet: M = 6.8*10^22 kg

(a)   The size of the acceleration due to gravity on the surface of Planet X,

gx = G*M / R^2 = (6.67*10^-11)*(6.8*10^22) / (2*10^6)^2 = 1.134 m/s^2

(b)    The weight of a 10 kg mass on the surface of Planet X,

W = m*gx = 10*(1.134) = 11.34 N

(c) For the time i would take for a ball dropped from a height of 4 m to hit the ground,

Using: h = (1/2)*gx*t^2

=> t = sqrt[ (2*h) / gx ] = sqrt[ (2*4) / 1.134 ] = 2.66 s

(d) For the value of gx at 3 of Planet X's radii above the planet's surface,

Distance from planet's surface = 3R

Therefore, distance from planet's center = R + 3R = 4R

gx = G*M / (4R)^2 = [G*M / R^2]*(1/16)

As calculated in part a, G*M / R^2 = 1.134 m/s^2

Therefore, here gx = (1.134)*(1/16) = 0.0709 m/s^2

(e) Let orbital speed = v

=> Centripetal acceleration = v^2/d

Also, centripetal acceleration = GMs / d^2, where Ms = Sun's mass

=> v^2 / d = GMs / d^2

=> v = sqrt[ GMs / d ]

= sqrt[ (6.67*10^-11)*(1.989*10^30) / (3.6*10^12) ] = 6070.56 m/s

=> Centripetal acceleration = v^2 / d  = (6070.56)^2 / (3.6*10^12) = 1.024*10^-5 m/s^2

(f) Circumference of X's orbital = 2**d

= 2**(3.6*10^12) = 2.262*10^13 m

One year = Time taken to make one revolution

  = Circumference of orbital / Orbital speed

= (2.262*10^13) / 6070.56 = 3.726*10^9 s

On Earth, One year = 365 days = 365*24 hours = 365*24*3600 s = 3.154*10^7 s

Therefore, One year on X/ One year on Earth

= (3.726*10^9) / (3.154*10^7) =  118 Earth Years

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