Home
Class 11
PHYSICS
A satellite is in an elliptic orbit arou...

A satellite is in an elliptic orbit around the earth with aphelion of 6R and perihelion of 2R where R = 6400 km is the radius of the earth . Find the velcoity of the satellite at apogee and perigee.[ `G = 6.67 xx10^(-1)` SI unit and `M = 6 xx 10^(24)kg` ]

Text Solution

Verified by Experts

`implies` Given , `r_p` = radius of perihelion = 2R
`r_a` = radius of aphelion = 6R
Hence , we can write,
`r_(a) = a(1+e) = 6R " " ...(i)`
`r_p = a (1-e) = 2 R " " ...(ii)`
Solving equ (i) and (ii) ,we get
eccentricity , `e =1/2`
Angular momentum remains unchanged
`:. mv_pr_p = mv_ar_a`
`:. (v_a)/(v_p) = r_p/r_a = (2R)/(6R) =1/3`
Energy is same at perigee and apogee,
`1/2mv_p^2-(GMm)/r_p=1/2mv_a^2-(GMm)/r_a`
`:. v_p^2(1-1/9)=-2GM(1/r_a-1/r_p)=2GM(1/r_p-1/r_a)`
(By putting `v_a=v_p/3`)
`v_p=([2GM(1/r_p-1/r_a)]^(1/2))/([1-(v_a/v_p)^2]^(1/2))=[((2GM)/R(1/2-1/6))/((1-1/9))]^(1/2)`
`=((3/2GM)/(8/9R))^(1/2)=sqrt(3/4(GM)/R) ` = 6.85 km/s
`v_p = 6.85km//s`
For circular orbit of radius r,
`v_c` = orbita velocity = `sqrt((GM)/r)`
For `r = 6R ,v_c =sqrt((GM)/(6R)) = 3.23 km//s`
Hence , to increase to a circular orbit at apogee, we have to increase the velocity by (3.22-2.28) = 0.95 km/s . This can be done by suitably firing rockets from the satellite .
Promotional Banner

Topper's Solved these Questions

  • GRAVITATION

    KUMAR PRAKASHAN|Exercise Section - E Multiple Choice Questions (MCQs)|73 Videos

  • GRAVITATION

    KUMAR PRAKASHAN|Exercise Section - F Questions from Module|20 Videos

  • GRAVITATION

    KUMAR PRAKASHAN|Exercise Section - D NCERT Exemplar Solutions ( Short Answer Type Question )|6 Videos

  • LAW OF MOTION

    KUMAR PRAKASHAN|Exercise (QUESTION PAPER) SECTION-D|1 Videos

Similar Questions

Explore conceptually related problems

A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth, where R is the radius of the earth. The time period of another satellite at a height of 2.5 R from the surface of the earth is …… hours.

A geostationary satellite is orbiting the earth at a height of 5R above the surface of the earth, 2R being the radius of the earth. The time period of another satellite in hours at a height of 2R form the surface of the earth is

A satellite is moving around the earth's with speed v in a circular orbit of radius r . If the orbit radius is decreases by 1% , its speed will

A satellite of mass m revolves around the earth of radius R at a height x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is

As you have learnt in the text a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth's gravity at the site of this satellite ? (Take the potential energy at infinity to be zero). Mass of the earth 6.0xx 10^(24(kg) , Radius = 6400 km .

An artificial satellite is moving in a circular orbit around the earth with a speed of equal to half the magnitude of escape velocity from earth. (i). Determine the height of the satellite above the earth's surface (ii). If the satellite is stopped suddenly in its orbit and allowed to fall freely on the earth. Find the speed with it hits and surface of earth. Given M="mass of earth & R "="Radius of earth"

A satellite of mass m revolves around the earth of radius R at a height x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is .........

An artificial satellite is moving in a circular orbit around the earth with a speed of equal to half the magnitude of escape velocity from earth. If the satellite is stopped suddenly in its orbit and allowed to fall freely on the earth. Find the speed with which it hits the surface of earth. Given, M=mass of earth & R =Radius of earth

A satellite A of mass m is at a distance of r from the surface of the earth. Another satellite B of mass 2m is at a distance of 2r from the earth's centre. Their time periods are in the ratio of

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).