The potential energy of a satellite is given as PE =λ( KE ) where, P E= potential energy of the satellite and KE = kinetic energy of the satellite. The value of the constant λ is

Question

The potential energy of a satellite is given as PE =λ( KE ) where, P E= potential energy of the satellite and KE = kinetic energy of the satellite. The value of the constant λ is (A) -2 (B) 2 (C) -1 / 2 (D) +1 / 2

Tardigrade Admin · Accepted Answer

PE of satellite =(-G m ME/(RE+h)) ..... (i) KE of satellite =+(1/2) (G m ME/(RE+h)) .....(ii) From Eqs. (i) and (ii), we get PE =-2 KE ⇒ λ =-2

Q. The potential energy of a satellite is given as
$PE = λ (K E)$
where, $PE =$ potential energy of the satellite
and $K E =$ kinetic energy of the satellite.
The value of the constant $λ$ is

$- 2$

$2$

$- 1/2$

$+ 1/2$

PE of satellite $= \frac{- G m M _{E}}{( R _{E} + h )}$ ..... (i)
KE of satellite $= + \frac{1}{2} \frac{G m M _{E}}{( R _{E} + h )}$ .....(ii)
From Eqs. (i) and (ii), we get
$PE = - 2 K E$
$\Rightarrow λ = - 2$

Q. The potential energy of a satellite is given as PE=λ(KE) where, PE= potential energy of the satellite and KE= kinetic energy of the satellite. The value of the constant λ is

−2

2

−1/2

+1/2

PE of satellite =(RE​+h)−GmME​​ ..... (i) KE of satellite =+21​(RE​+h)GmME​​ .....(ii) From Eqs. (i) and (ii), we get PE=−2KE ⇒λ=−2

Q. The potential energy of a satellite is given as
$PE = λ (K E)$
where, $PE =$ potential energy of the satellite
and $K E =$ kinetic energy of the satellite.
The value of the constant $λ$ is

$- 2$

$2$

$- 1/2$

$+ 1/2$

PE of satellite $= \frac{- G m M _{E}}{( R _{E} + h )}$ ..... (i)
KE of satellite $= + \frac{1}{2} \frac{G m M _{E}}{( R _{E} + h )}$ .....(ii)
From Eqs. (i) and (ii), we get
$PE = - 2 K E$
$\Rightarrow λ = - 2$