Question

1. A 9.4×1021 kg moon orbits a distant planet in a circular orbit of radius 1.5×108 m. It experiences a 1.1×1019 N gravitational pull from the planet. What is the moon’s orbital period in earth days?

Answer 1

Answer:

26days

Explanation:

Centripetal force is the force that tend to pull an object moving in a circular path towards the centre.

Mathematically;

Centripetal force Fc = mv²/r where;

m is the mass of the moon = 9.4×10^21kg

v is the velocity of the moon in a circular orbit

r is the radius of the circular orbit = 1.5×10^8m

Fc is the gravitational force experienced by the moon = 1.1×10^19N

Before we can get the period, we need to know the velocity at which the moon is moving around the circular orbit.

Substituting the given datas in the formula above we have;

1.1×10^19 = 9.4×10^21v²/1.5×10^8

Cross multiplying we will have;

1.65×10^27 = 9.4×10^21v²

v² =1.65×10^27/9.4×10^21

v² = 0.176×10^6

v= √0.176×10^6

v = 419.53m/s

To get the moon orbital period T, we will use the relationship;

T = 2πr/v where

T = 2π(1.5×10^8)/419.53

T = 9.42×10^8/419.53

T = 2,246,508.7seconds

Converting to days

If 24hours = 1day

(24×60×60)seconds = 1day

If 86400seconds = 1day

2,246,508.7seconds = x days

Cross multiplying we have

x × 86400 = 2,246,508.7

x = 2,246,508.7/86,400

x = 26days

Therefore the moon orbital period in earth days is approximately 26days.

Answer 2

Answer:

26 days

Explanation:

m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

We know Fc = $\frac{m v^{2} }{r}$

==> 1.1 × $10^{19}$ = (9.4 × $10^{21}$ × $v^{2}$ ) ÷ 1.5 × $10^{8}$

==> 1.1 × $10^{19}$ = $v^{2}$ × 6.26×$10^{13}$

==> $v^{2}$ =  1.1 × $10^{19}$ ÷ 6.26×$10^{13}$

==> $v^{2}$ = 0.17571885 × $10^{6}$

==> v= 0.419188323 × $10^{3}$ m/sec

==> v= 419.188322834 m/s

Putting value of r and v from above in ;

T= 2πr ÷ v

==> T= 2×3.14×1.5×$10^{8}$ ÷ 0.419188323 × $10^{3}$

==> T = 22.472× 100000 = 2247200 sec

but

86400 sec = 1 day

==> 2247200 sec= 2247200 ÷ 86400 = 26 days