A satellite with a mass of 290 kg moves in a circular orbit 4.00 ? 10 7 m above the...


A satellite with a mass of 290 kg moves in a circular orbit {eq}4.00 \ * \ 10^{7} \ m {/eq} above the Earth's surface.

a. What is the gravitational force on the satellite?

b. What is the speed of the satellite?

c. What is the period of the satellite?

Earth's Pull

The gravitation force existing between the Earth and any other object under its influence of gravity field is analogous to the coulombic force of attraction between two charges. The force of gravitation is directly proportional to the product of their masses in ratio with the square of the distance between them.

Answer and Explanation: 1

(a.) The force on the satellite due to Earth gravitation pull will be computed from the centre of the Earth. Therefore, the given data is,

{eq}r= (4\times 10^7)+(0.6378\times 10^7) {/eq}

= {eq}4.64\times 10^7 m {/eq} (second term is the radius of earth)

{eq}m_e {/eq} = mass of the Earth ({eq}5.972\times 10^{24} {/eq}kg)

{eq}m_s {/eq} = mass of the satellite (290 kg)

{eq}G = 6.67\times 10^{11} {/eq}{eq}Nm^2/kg^2 {/eq}

Plug in the given values in the expression for force of gravitation, {eq}F_g= G\dfrac{m_e m_s}{r^2} {/eq}

{eq}\Rightarrow F_g= (6.67\times 10^{11})\dfrac{(5.972\times 10^{24})(290)}{(4.64\times 10^{7})}\\ {/eq}

{eq}2.49\times 10^{31} kg.m/s^2 {/eq} (towards the centre of the earth)

The above force has a special name called centripetal acceleration which is the cause for circular motion.

(b.) Speed of the satellite would be the circular velocity given by the mathematical expression, {eq}v_s= \sqrt{\dfrac{G\times m_e}{R}} {/eq}

Plug in the values we get,

{eq}\Rightarrow v_s= \sqrt{\dfrac{6.67\times 10^{11}\times5.972\times 10^{24}}{4\times 10^7}}\\ \Rightarrow v_s = \sqrt{9.95\times 10^{28}}\\ {/eq}

= {eq}3.15\times 10^{14} m/s {/eq}

(c.) The mathematical expression for the period of the satellite is,

{eq}T=2 \pi\sqrt{\dfrac{r^3}{Gm_e}} {/eq}

Plug in the values we get,

{eq}\Rightarrow T=2 \pi\sqrt{\dfrac{r^3}{Gm_e}}\\ \Rightarrow T=2(3.14)\sqrt{\dfrac{(4\times 10^7)^3}{(6.67\times 10^{11})(5.972\times 10^{24})}}\\ \Rightarrow T=(6.28)\sqrt{\dfrac{64\times 10^{21}}{39.8\times 10^{35}}}\\ \Rightarrow T= (6.28)\sqrt{1.608\times 10^{-14}}\\ \Rightarrow T = (6.28) (1.27\times 10^{-7})\\ {/eq}

{eq}7.97 \times 10^{-7} {/eq} seconds.

Learn more about this topic:

Gravitational Pull of the Earth: Definition & Overview


Chapter 15 / Lesson 17

Earth's gravitational pull is often misunderstood, but without it, life on Earth would be impossible. In this lesson, we'll define the gravitational pull and give some examples of how it is used. A quiz is provided to test your understanding.

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