Question

A Molniya orbit is a highly eccentric orbit of a communication satellite so as to provide continuous communications coverage for Scandinavian countries and adjacent Russia. The orbit is positioned so that these countries have the satellite in view for extended periods in time (see below). If a satellite in such an orbit has an apogee at 48,000.0 km as measured from the center of Earth and a velocity at apogee of 3.7 km/s, what would be its velocity (in km/s) at perigee measured at 740.0 km altitude? (Enter the magnitude.)

Answer 1

Answer: 18.3 km/s

Step-by-step explanation:

If a satellite in Molniya orbit has an apogee at 48.000 km as measured from the center of Earth, and a velocity of 3.7 km/s. Its velocity in at perigee would be 18.3 km/s.

Answer 2

The velocity of the satellite at the perigee is
`v_(p) = 24.97km/s`.

Step-by-step explanation:

The velocity of the satellite at the perigee can be found by means of the angular momentum:

`L = mrv` (1)

Since there is no torque acting on the system, it can be express in the following way:

`t = (\Delta L)/(\Delta t)`

`t \Delta t = \Delta L`

`\Delta L = 0`

`L_(a) - L_(p) = 0`

`L_(a) = L_(p)` (2)

Replacing equation 1 in equation 2 it is gotten:

`mr_(a)v_(a) =mr_(p)v_(p)` (3)

Where m is the mass of the comet,
`r_(a)`is the orbital radius at the apogee,
`v_(a)` is the speed at the apogee,
`r_(p)` is the orbital radius at the perigee and
`v_(p)` is the speed at the perigee.

From equation 3
`v_(p)` will be isolated:

`v_(p) = (mr_(a)v_(a))/(mr_(p))`

`v_(p) = (r_(a)v_(a))/(r_(p))` (4)

Before using equation 4 it is necessary to found the orbital radius of the satellite at the perigee, that can be determined from the sum of the Earth's radius and the height of the satellite above the surface at this point (740.0 km).

`r_(p) = 6371km+740.0km`

`r_(p) = 7111km`

Finally, equation 4 can be used:

`v_(p) = ((48000.0 km)(3.7 km/s))/((7111 km))`

`v_(p) = 24.97km/s`

Hence, the velocity of the satellite at the perigee is
`v_(p) = 24.97km/s`.