Answer:
a.
[tex]F=G\cdot\dfrac{M \cdot m}{r^{2}}[/tex]
b.
[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}}[/tex]
c.
[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}} \approx 1.144 \times 10^{13} \ N[/tex]
d. The force of gravity acting on the satellite is approximately 1.144 × 10¹³ N
Explanation:
a. The formula for finding the force of gravity, F, acting object on an object is given as follows;
[tex]F=G\cdot\dfrac{M \cdot m}{r^{2}}[/tex]
Where;
F = The force acting between the Earth and the object
G = The gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
M = The mass of the Earth = 5.97 × 10²⁴ kg
m = The mass of the object
r = The distance between the center of the Earth and the object
b. Finding the gravitational force, 'F', between the Earth and the scientific satellite, we have;
The given mass of the satellite, m = 1,300 kg
The distance between the center of the Earth and the center of the satellite, r = The length of the radius of the Earth + The height of orbit of the satellite
The given height of orbit of the satellite, h = 200 km
∴ r = R + h = 6,378 km + 200 km = 6,578 m
Therefore, by plugging in the values, we get;
[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}}[/tex]
c. Solving the above equation gives;
[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}} \approx 1.144 \times 10^{13} \ N[/tex]
d. The force of gravity acting on the satellite, F ≈ 1.144 × 10¹³ Newton
