PHYSICS MIDDLE SCHOOL

An artificial satellite circling the Earth completes each orbit in 129 min(a) Find the altitude of the satellite.

(b) What is the value of g at the location of this satellite?

Answers

Answer 1

Answer:

(a) $2.09\cdot 10^6 m$ above Earth's surface

The orbital speed of a satellite orbiting the Earth can be found using the equation

$v=\sqrt{(GM)/(r)}$

where

G is the gravitational constant

$M=5.98\cdot 10^(24) kg$ is the Earth's mass

r is the radius of the satellite's orbit

The orbital speed can also be rewritten as the ratio between the circumference of the orbit and the orbital period, T:

$(2\pi r)/(T)$

where

T = 129 min = 7740 s is the period

Combining the two equations,

$(2\pi r)/(T)=\sqrt{(GM)/(r)}$

And solving for r,

$((2\pi)^2 r^2)/(T^2)=(GM)/(r)\nr=\sqrt[3]{(GMT^2)/((2\pi)^2)}=\sqrt[3]{((6.67\cdot 10^(-11))(5.98\cdot 10^(24))(7740)^2)/((2\pi)^2)}=8.46\cdot 10^6 m$

This is, however, the orbital radius: this means we have to subtract the Earth's radius to find the altitude of the satellite, which is

$R=6.37\cdot 10^6 m$

therefore, the altitude of the satellite is

$h=r-R=8.46\cdot 10^6 - 6.37\cdot 10^6 =2.09\cdot 10^6 m$

b) $5.57 m/s^2$

The value of g at the location of the satellite is given by

$g=(GM)/(r^2)$

where:

G is the gravitational constant

$M=5.98\cdot 10^(24) kg$ is the Earth's mass

$r=8.46 \cdot 10^6 m$ is the radius of the satellite's orbit

Substituting into the equation, we find

$g=((6.67\cdot 10^(-11))(5.98\cdot 10^(24)))/((8.46\cdot 10^6)^2)=5.57 m/s^2$

Answer 2

Answer:

Final answer:

The satellite orbits at an altitude of approximately 800 km. The gravitational constant, 'g', at this location is approximately 8.66 m/s^2.

Explanation:

The orbital period of an artificial satellite can be used to calculate the altitude at which it orbits. For a satellite that completes each orbit in 129 min (or approximately 2.15 hr), we can apply Kepler's third law which states that the square of the period of a satellite is proportional to the cube of its semi-major axis (distance from the center of the Earth to the satellite).

The formula for the altitude is given by: h = [(GMT^2)/(4π^2)]^(1/3) - R, where G is the gravitational constant, M the mass of Earth, T the orbital period, and R the Earth's radius. With the values G=6.67 x 10^-11 N(m/kg)^2, M=5.98 x 10^24 kg, T=2.15 hr = 7740s, and R=6.371 x 10^6 m, we get h approximately equals 800 km.

The value of 'g' at the satellite's location is given by g = GM/(R+h)^2. Substituting the aforementioned values, we get g to be approximately 8.66 m/s^2. This is less than the 9.81 m/s^2 at Earth's surface due to the increased distance from the Earth's center.

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