Question

The radius of the Earth’s orbit around the sun (assumed to be circular) is 1.50∙108 km, and the Earth travels

around this obit in 365 days. The mass of the Earth is 5.98·1024 kg, and mass of the Sun is 1.99∙1030 kg.

(a) What is the magnitude of the orbital velocity of the Earth, in m/s?

(b) What is the radial acceleration of the earth toward the sun, in m/s2?

(c) What is the magnitude of centripetal force acting on the Earth?

(d) What is responsible for providing this centripetal force?

(e) Calculate the gravitational acceleration OF the Earth (not ON the Earth). Hint: think of your answer to

part (d), and set two forces equal to each other.

(f) What would happen to the Earth’s motion if gravity was turned off? Answer as precisely as possible.

Answer 1

a) orbital velocity,

vo=((G*mass of sun) /radius of orbit) ^0.5

vo=((6.67*10^-11*1.99*10^30)/1.50*10^8)^0.5

vo=9.4*10^5 m/s

b) radial acceleration,

ar=(vo^2)/r

ar=(9.4*9.4*10^10)/1.50*10^8

ar=5.89*10^3 m/s^2

c)centripetal force,

Fc=mass of earth*ar

Fc=5.98*10^24*5.89*10^3

Fc=35.22*10^27 N

d) When an object moves on a circle its feels a force known as centripetal force to maintain its circular path and to be in its orbit. The sun and the earth have gravitational force between them and to maintain balance this centripetal force comes into existence.

E) here gravitational acceleration means gravitational constant, it can be calculated by putting gravitational force and centripetal force be equal,

So, (G*ms*me) /r*r=(me*vo*vo)/r

G=(vo*vo*r)/ms

G=(9.4*9.4*10^10*1.50*10^8)/(1.99*10^30)

G=6.660*10^(-11)Nm^2kg^-2

f) If there will be no gravity then there will be no gravitational force between sun and earth and any other planet in the atmosphere which result in the random movement of the earth. There will be no particular motion of earth as it has now a circular orbit around sun.