Answer:
the gravitational force between the satellite and the planet is 32.4 * 10³N
Explanation:
It is given that,
Mass of the satellite, m = 6500 kg
Speed of the satellite, v = 6.7 × 10³ m/s
distance to the center of the planet = 9 × 10⁶m
Let F is the magnitude of the gravitational force exerted on the satellite by the planet. The centripetal force is equal to the gravitational force. It is equal to :
[tex]F = \frac{mv^2}{R}[/tex]
[tex]F = \frac{6500 \times (6.7 * 10^3)}{9 * 10^6}[/tex]
= 32.4 * 10³N
the gravitational force between the satellite and the planet is 32.4 * 10³N
Complete Question
The complete question is shown on the first uploaded image
Answer:
Option D is the Answer
Explanation:
The formula for the velocity of the satellite is
[tex]v =\sqrt{\frac{GM}{R} }[/tex]
R is the distance to the center o the planet = [tex]9*10^6m[/tex]
G is the gravitational constant which is equal to [tex]= 6.67*10^{-11} \frac{Nm^2}{kg^2}[/tex]
M is the mass of the planet
Making M the subject of the formula we have
[tex]M = \frac{(v^2)(R)}{G}[/tex]
[tex]M = \frac{(6.7*10^3)^2(9*10^6)}{6.67*10^{-11}} = 60.57*10^{23}Kg[/tex]
The formula for the gravitational force between two object
[tex]F = \frac{GMm}{R^2}[/tex]
Where
[tex]= \frac{6.67*10^{-11}(6.057*10^{24})(6500)}{(9*10^6)^2}[/tex]
[tex]F = 32,419.90 N[/tex]
[tex]= 3.24*10^{4} N[/tex]
