Question

The period of a satellite circling planet Nutron is observed to be 84 s (seconds) when it is in circular orbit with a radius of 8.0 x 106 m. What is the mass of planet Nutron?
a. 6.2 x 1028 kg
b. 5.0 x 1028 kg
c. 5.5 x 1028 kg
d. 4.3 x 1028 kg
e. 3.7 x 1028 kg

Answer 1

Answer:

d. 4.3 x 10²⁸ kg.

Explanation:

• The force that keeps the satellite circling the planet, is the centripetal force, which can be expressed as follows:

       $F_{c} = m_{s}* \frac{v^{2}}{r} (1)$

• This force is not a different force, is just the net force acting on the satellite towards the planet, which is simply the gravitational force between the planet and satellite masses, that we can write as follows, according the Newton's Universal Law of Gravitation:

       $F_{g} = G*\frac{m_{s}*m_{p} }{r^{2}} (2)$

• From (1) and (2) as they are equal each other we could get the mass of the planet.
• However, we don't know the value of the linear speed v, but we know the value of the period T instead.
• By definition, T is the time needed for the satellite to complete a single circumference of the orbit, so, applying the definition of average velocity, we get:

       $T = \frac{\Delta s }{v} = \frac{2*\pi *r}{v} (3)$

• Solving (3) for v, replacing the value in (1), and solving finally for mp in (2), we get:
• $m_{p} = \frac{4*\pi ^{2} *r^{3} }{T^{2}*G} = 4.3e28 kg (4)$

Answer 2

The mass of planet Nutron is ; ( D ) $M_{b} = 4.3 * 10^{28} kg$

Given data :

Satellite period of circling ( T )  = 84 secs

Radius ( r ) = 8.0 * 10⁶ m

Determine the mass of the planet Nutron

we will apply the equation representing the centripetal force

Centripetal force ( Fc ) = $m_{s} * \frac{v^2}{r} ----- ( 1 )$

applying Newton's universal law of Gravitation to equation ( 1 )

Fg = $G * \frac{m_{s} * m_{b} }{r^2}$   ------ ( 2 ).

note : equation ( 1 ) equals equation ( 2 )

First step : express T in terms of v and change in position ( Δs )

Given that v ( linear speed ) is unknown

T = Δs / v = $\frac{2\pi r}{v}$   --- ( 3 )

∴ v = [ 2 * π * ( 8 * 10⁶ ) ] /  ( 84 )

     = 598639.45 m/s

Input result in equation ( 1 ) and solve for the mass of planet Nutron in equation 2

∴  Mass of planet Nutron = $M_{b} = 4.3 * 10^{28} kg$

Hence we can conclude that The mass of planet Nutron is ; ( D ) $M_{b} = 4.3 * 10^{28} kg$

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