A 2150 kg satellite used in a cellular telephone network is in a circular orbit at a height of 780 km above the surface of the earth. (a) What is the gravitational force on the satellite? (b) What fraction is this force of the satellite’s weight at the surface of the earth?

Where [tex]M=5.97\times10^{24}kg[/tex] is Earth's mass, [tex]m=2150kg[/tex] is the satellite mass, [tex]r=R+h[/tex] is the distance between their centers, where [tex]h=780000m[/tex] is the height of the satellite (from Earth's surface) and [tex]R=6371000m[/tex] is Earth's radius, and [tex]G=6.67\times10^{-11}Nm^2/kg^2[/tex] is the gravitational constant.

a) With these values we then have:

[tex]F=\frac{GMm}{r^2}=\frac{(6.67\times10^{-11}Nm^2/kg^2)(5.97\times10^{24}kg)(2150kg)}{(6371000m+780000m)^2}=16741.9N[/tex]

b) And the fraction this force is of the satellite’s weight W=mg is:

[tex]\frac{F}{W}=\frac{GMm}{mgr^2}=\frac{GM}{gr^2}=\frac{(6.67\times10^{-11}Nm^2/kg^2)(5.97\times10^{24}kg)}{(9.8m/s^2)(6371000m+780000m)^2}=0.795[/tex]

hamzaahmeds hamzaahmeds · Answer 2 · 2021-10-16T10:24:19+02:00

This question involves the concept of Newton's Law of Gravitation, Gravitational Force, and Weight.

(a) The Gravitational Force on satellite is "1.682 x 10⁴ N (OR) 16.82 KN".

(b) This Gravitational Force is "0.8 times" the satellite's weight at the surface of the earth.

(a)

We will use the formula of Newton's Gravitational Law here to find out the Gravitational Force:

[tex]F = \frac{Gm_1m_2}{r^2}[/tex]

where,

F = Gravitational Force = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

m₁ = mass of earth = 6 x 10²⁴ kg

m₂ = mass of satellite = 2150 kg

r = distance between center of earth and the center of the satellite

r = distance of satellite from the surface of earth + radius if earth

r = 780000 km + 6371000 m = 7151000 m

Therefore,

[tex]F = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)(2150\ kg)}{(7151000\ m)^2}[/tex]

F = 1.682 x 10⁴ N = 16.82 KN

(b)

Now we will divide this force by the weight of the satellite on the surface of the earth:

[tex]\frac{F}{Weight} = \frac{1.682\ x\ 10^4\ N}{mg}\\\\\frac{F}{W} = \frac{1.682\ x\ 10^4\ N}{(2150\ kg)(9.81\ m/s^2)}\\\\[/tex]

F = 0.8 W

Learn more about Newton's Law of Gravitation here:

brainly.com/question/17931361?referrer=searchResults

The attached picture illustrates Newton's Law of Gravitation.