Question #26253

Obtain an expression for the time period of a satellite orbiting the earth. A space shuttle is in a

circular orbit at a height of 250 km from the earth’s surface, where the acceleration due to earth’s

gravity is 0.93 g. Calculate the period of its orbit. Take g = 9.8 ms−2 and the radius of the earth

R = 6.37 × 106m.

circular orbit at a height of 250 km from the earth’s surface, where the acceleration due to earth’s

gravity is 0.93 g. Calculate the period of its orbit. Take g = 9.8 ms−2 and the radius of the earth

R = 6.37 × 106m.

Expert's answer

Newton's second law:

F=ma, F - force of gravity, m - satellite's mass, a - acceleration of satellite.

According to the problem statement, F=mg1 and g1=0.93g, so F=0.93mg

Satellite moves in a circle with a constant angular velocity around the Earth, so a=w^2*(R+r) where w - angular velocity, R - radius of the Earth, r - height of satellite's orbit.

The result if:

0.93mg=mw^2*(R+r)

0.93g=w^2*(R+r)

It's obvious that w=2pi/T, so

0.93g=(2pi/T)^2*(R+r)

T=2pi*sqrt((R+r)/0.93g)

T=2*3.14*sqrt((6.37*10^6+0.25*10^6)/(0.93*9.8))=5.355*10^3(s)=1 hour 29 minutes and 15 seconds

F=ma, F - force of gravity, m - satellite's mass, a - acceleration of satellite.

According to the problem statement, F=mg1 and g1=0.93g, so F=0.93mg

Satellite moves in a circle with a constant angular velocity around the Earth, so a=w^2*(R+r) where w - angular velocity, R - radius of the Earth, r - height of satellite's orbit.

The result if:

0.93mg=mw^2*(R+r)

0.93g=w^2*(R+r)

It's obvious that w=2pi/T, so

0.93g=(2pi/T)^2*(R+r)

T=2pi*sqrt((R+r)/0.93g)

T=2*3.14*sqrt((6.37*10^6+0.25*10^6)/(0.93*9.8))=5.355*10^3(s)=1 hour 29 minutes and 15 seconds

## Comments

## Leave a comment