Answer:
A) F_g = 26284.48 N
B) v = 7404.18 m/s
C) E = 19.19 × 10^(10) J
Explanation:
We are given;
Mass of satellite; m = 3500 kg
Mass of the earth; M = 6 x 10²⁴ Kg
Earth circular orbit radius; R = 7.3 x 10⁶ m
A) Formula for the gravitational force is;
F_g = GmM/r²
Where G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²
Plugging in the relevant values, we have;
F_g = (6.67 × 10^(-11) × 3500 × 6 x 10²⁴)/(7.3 x 10⁶)²
F_g = 26284.48 N
B) From the momentum principle, we have that the gravitational force is equal to the centripetal force.
Thus;
GmM/r² = mv²/r
Making v th subject, we have;
v = √(GM/r)
Plugging in the relevant values;
v = √(6.67 × 10^(-11) × 6 x 10²⁴)/(7.3 x 10⁶))
v = 7404.18 m/s
C) From the energy principle, the minimum amount of work is given by;
E = (GmM/r) - ½mv²
Plugging in the relevant values;
E = [(6.67 × 10^(-11) × 3500 × 6 × 10²⁴)/(7.3 x 10⁶)] - (½ × 3500 × 7404.18)
E = 19.19 × 10^(10) J
Saying no and not throwing fits and manners.
Answer:
a
Explanation:
what the heck is a medium
the equation of the tangent line must be passed on a point A (a,b) and
perpendicular to the radius of the circle. <span>
I will take an example for a clear explanation:
let x² + y² = 4 is the equation of the circle,
its center is C(0,0). And we assume that the tangent line passes to the point
A(2.3).
</span>since the tangent passes to the A(2,3), the line must be perpendicular to the radius of the circle.
<span>Let's find the equation of the line parallel to the radius.</span>
<span>The line passes to the A(2,3) and C (0,0). y= ax+b is the standard form of the equation. AC(-2, -3) is a vector parallel to CM(x, y).</span>
det(AC, CM)= -2y +3x =0, is the equation of the line // to the radius.
let's find the equation of the line perpendicular to this previous line.
let M a point which lies on the line. so MA.AC=0 (scalar product),
it is (2-x, 3-y) . (-2, -3)= -4+4x + -9+3y=4x +3y -13=0 is the equation of tangent
