Question

The same 1710 kg artificial satellite is placed into circular orbit at the same altitude of 2.6x10° m around an exoplanet with the same radius as the Earth, but twice the mass. a. What is the orbital speed of the satellite? b. What is the period of the satellite? C. What is the kinetic energy of the satellite? d. What is the total energy of the satellite?

Answer 1

Given:

mass of satellite, m = 1710 kg

altitude, h =
`2.6* 10^(6) m`

G =
`6.67* 10^(-11)`

we know

mass of earth,
`M_(E)` =
`5.972* 10^(24) kg`

Here, according to question we will consider

`2M_(E)` =
`11.944* 10^(24) kg`

radius of earth,
`R_(E)` =
`6.371* 10^(6) m`

Formulae Used and replacing
`M_(E)` by
`2M_(E)` :

1).
`v = \sqrt{(2GM_(E))/(R_(E) + h)}`

2).
`T = \sqrt{(4\pi ^(2)(R_(E) + h)^(3))/(2GM_(E))}`

3).
`KE = (1)/(2)mv^(2)`

4).
`Total Energy, E = -(2GM_E* m)/(2(R_(E) + h))`

where,

v = orbital velocity of satellite

T = time period

KE = kinetic energy

Solution:

Now, Using Formula (1), for orbital velocity:

`v = \sqrt{(6.67 * 10^(-11) * 11.944 * 10^(24))/(6.371 * 10^(6) + 2.6 * 10^(6))`

v =
`9.423 * 10^(3)` m/s

Using Formula (2) for time period:

`T = \sqrt{(4\pi ^(2)(6.371* 10^(6) + 2* 10^(6))^(3))/(6.67* 10^(-11)* 9.44* 10^(24))}`

`T = 6.728* 10^(3) s`

Now, Using Formula(3) for kinetic energy:

`KE = (1)/(2)(9.44* 10^(24))(9.42* 10^(3))^(2)`

`KE = (1)/(2)(1710)(9.42* 10^(3))^(2) = 7.586* 10^(10) J`

Now, Using Formula(4) for Total energy:

`E = -(6.67* 10^(-11)* 9.44* 10^(24)* 1710)/(2( 6.371* 10^(6) + 2.6* 10^(6)))`

`E = - 7.59* 10^(10) J`