A 50-kg satellite circles planet Cruton every 5.6 h in an orbit with a radius of 12 * 106 m. What is the magnitude of the gravitational force on the satellite by planet Cruton?
A)63 N
B)58 N
C)68 N
D)73 N
E)50 N

Question

Quizplus · Accepted Answer

Using the formula for gravitational force:

$F = \frac{Gm_1m_2}{r^2}$

where 
- $G$ is the gravitational constant ($6.67	imes10^{-11}$ N$m^2/kg^2$), 
- $m_1$ and $m_2$ are the masses of the two objects, and 
- $r$ is the distance between their centers.

Converting the given values into SI units:

$m_1 =$ mass of planet Cruton $= ?$ (not given) 
$m_2 =$ mass of satellite $= 50$ kg 
$r =$ radius of satellite's orbit $= 12	imes10^6$ m

We can use Kepler's third law to find the mass of planet Cruton:

$T^2 = \frac{4\pi^2}{GM}r^3$

where 
- $T$ is the period of the satellite's orbit ($5.6$ hours $= 20,160$ seconds), 
- $G$ is the gravitational constant, 
- $M$ is the mass of the planet.

Solving for $M$:

$M = \frac{4\pi^2}{GT^2}r^3 = \frac{4\pi^2}{(6.67	imes10^{-11})(20,160)^2}(12	imes10^6)^3 = 6.12	imes10^{24}$ kg

Now we can calculate the gravitational force:

$F = \frac{Gm_1m_2}{r^2} = \frac{(6.67	imes10^{-11})(6.12	imes10^{24})(50)}{(12	imes10^6)^2} \approx 58	ext{ N}$

Therefore, the answer is B) 58 N.

A 50-Kg Satellite Circles Planet Cruton Every 5