Home
Class 12
PHYSICS
A satellite is a taken to a height equal...

A satellite is a taken to a height equal to the radius of the Earth and then projected horizontally with a speed of 7km/s . State the nature of its orbit.

Text Solution

Verified by Experts

Data : `h= R_(E), G = 6.67 xx 10^(-11) N.m^(2)//kg^(2), M_(E) = 6xx 10^(24)` kg,
` R_(E) = 6.4 xx 10^(6)` m
Critical speed at an altitude h =` R_(E)`is
`v_(c)=sqrt((GM_(E))/(R_(E)+h))=sqrt((Gm_(E))/(2R_(E)))`
` = sqrt((6.67 xx 10^(11))(6xx10^(24)))/(2(6.4 xx 10^(6)))= sqrt((200.1 xx 10^(6))/(6.4))`
= `5.591 xx 10^(3) m//s = 5.591 ` km/s
And the escape speed from an orbit at that altitude,
` v_(c)=sqrt(2)v_(c)`
= 1.414 `xx` 5.591= 7.905 km/s
Since 7.905 km/s gt 7 km /s gt 5.591 km/s
` v_(e) gt v gt v_(c)`
Teh satellite's orbit will be elliptical.
Promotional Banner

Topper's Solved these Questions

  • SOLVED PROBLEMS - I

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise ROTATIONAL MOTION|26 Videos
  • SOLVED PROBLEMS - I

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise ELASTICITY|26 Videos
  • SOLVED PROBLEMS - I

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise ASSIGNMENTS|24 Videos
  • SHORT ANSWER QUESTIONS

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise Assignments|3 Videos
  • SOLVED PROBLEMS-II

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise Assignments|5 Videos

Similar Questions

Explore conceptually related problems

A satellite of mass m is orbiting around the earth at a height equal to twice the radius of the earth (R). Its potential energy is given by

A satellite is orbiting the earth in a circular orbit of radius r . Its

A satelite is revolving in a circular orbit at a height h above the surface of the earth of radius R. The speed of the satellite in its orbit is one-fourth the escape velocity from the surface of the earth. The relation between h and R is

A satellite is revolving in an orbit a height 20200 km from the suface of the earth . Determine the speed with which the satellite is revolving around the earth. Also calcutate the time taken by the satellite to compete on revolution around the earth.

A geostationary satellite is orbiting the Earth at a height of 6 R above the surface of Earth, where R is the radius of the Earth. The time period of another satellites is 6 sqrt(2) h . Find its height from the surface of Earth.

A satellite of mass m is in an elliptical orbit around the earth of mass M(M gt gt m) the speed of the satellite at its nearest point ot the earth (perigee) is sqrt((6GM)/(5R)) where R= its closest distance to the earth it is desired to transfer this satellite into a circular orbit around the earth of radius equal its largest distance from the earth. Find the increase in its speed to be imparted at the apogee (farthest point on the elliptical orbit).

A satellite is revolving around the earth. Ratio of its orbital speed and escape speed will be.