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An artificial satellite circling the Earth completes each orbit in 125 minutes.

An artificial satellite circling the Earth completes each orbit in 125 minutes. (a) Find the altitude of the satellite.(b) What is the value of g at the location of this satellite?

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animegirl02

5,0(70 marks)

(a) Altitude = 1.95 x 10⁶ m = 1950 km

(b) g = 5.9 m/s²

Explanation:

(a)

The time period of the satellite is given by the following formula:

$T^2 = \frac{4\pi^2r^3}{GM_E}$

where,

T = Time period = (125 min)( $\frac{60\ s}{1\ min}$ ) = 7500 s

r = distance of satellite from the center of earth = ?

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

M_E = Mass of Earth = 6 x 10²⁴ kg

Therefore,

$(7500\ s)^2 = \frac{4\pi^2r^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}\\\\r^3 = \frac{(7500\ s)^2(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}{4\pi^2}\\\\r = \sqrt[3]{5.7\ x\ 10^{20}\ m^3} \\$

r = 8.29 x 10⁶ m

Hence, the altitude of the satellite will be:

$Altitude = r - radius\ of\ Earth\\Altitude = 8.29\ x\ 10^6\ m - 6.34\ x\ 10^6\ m$

Altitude = 1.95 x 10⁶ m = 1950 km

(b)

The weight of the satellite will be equal to the gravitational force between satellite and Earth:

$Weight = Gravitational\ Force\\\\M_sg = \frac{GM_EM_s}{r^2}\\\\g = \frac{GM_E}{r^2}\\\\g = \frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(6\ x\ 10^{24}\ kg)}{(8.23\ x\ 10^6\ m)^2}$

g = 5.9 m/s²

elizabethburkha

4,5(1 marks)

idfk mate

Explanation:

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