Question

Two satellites are in circular orbits around a planet that has radius 9.00×106m. One satellite has mass 68.0 kg, orbital radius 6.00×107 m , and orbital speed 4800 m/sm/s. The second satellite has mass 84.0 kg and orbital radius 3.00×107 m.What is the orbital speed of this second satellite?v=?

Answer 1

Answer:

The orbital speed of the second satellite is 6788.29 m/s

Explanation:

Given;

Radius of the planet, R = 9.00 × 10⁶ m

Mass of the first, m₁ = 68.0 kg

Speed of the first planet, v₁ = 4800 m/s

Radius of the first planet, r₁ = 6.00 x 10⁷ m

Mass of the second planet, m₂ = 84.0 kg

Radius of the second planet, r₂ = 3.00 × 10⁷ m

Let the speed of the second planet, = v₂

From Newton's gravitational law;

$F = \frac{GM_pm}{r^2}$ -------equation (i)

where;

G is universal gravitational constant, G = 6.67 x 10⁻¹¹ N.m²/kg²

$M_p$ is mass of the planet

m is mass of the satellite

r is radius of the satellite

From Newton's second law of motion;

$F = ma=m\frac{v^2}{r}$ -----------equation (ii)

Equate (i) and (ii)

$\frac{GM_pm}{r^2} =\frac{mv^2}{r} \\\\\frac{GM_p}{r} =v^2\\\\v = \sqrt{\frac{GM_p}{r} }$

v is velocity of the satellite

r is radius of the satellite

Now, determine $M_p$, mass of the planet

$v_1 =\sqrt{\frac{GM_p}{r_1} } \\\\v_1^2 = \frac{GM_p}{r_1}\\\\M_p = \frac{v_1^2r_1}{G} = \frac{(4800)^2(6*10^7)}{6.67*10^{-11}}=2.0726*10^{25} \ kg$

Finally, determine the speed of the second satellite;

$v_2=\sqrt{\frac{GM_p}{r_2} }\\\\v_2=\sqrt{\frac{6.67*10^{-11}(2.0726*10^{25})}{3*10^7} }\\\\v_2 = 6788.29 \ m/s$

Therefore, the orbital speed of the second satellite is 6788.29 m/s

Answer 2

Answer: 6782 m/s

Explanation:

Given

Radius of the planet, r = 9*10^6 m

Mass of satellite 1, m1 = 68 kg

Radius of satellite 1, r1 = 6*10^7 m

Orbital speed of satellite 1, vs1 = 4800 m/s

Mass of satellite 2, m2 = 84 kg

Radius of satellite 2, r2 = 3*10^7 m

Orbital speed of satellite 2, vs2 = ?

We know that magnitude of gravitational force, F = (G.m.m•) / r²

Where,

m = mass of satellite

m• = mass of planet

r = radius of orbit

If we consider Newton's second law that states that, F = ma, thus

F(g) = ma(rad)

Where, a(rad) = v²/r

F(g) = mv²/r

Substituting in the initial equation

mv²/r = (G.m.m•) / r²

v² = (G.m•) / r

v = √[G.m•/r]

To find vs2, we first need to find mass of the planet, m• we know that G is a gravitational constant, so we plug in the values

vs1 = √[G.m•/r1]

4800 = √[(6.67*10^-11 * m•) / 6*10^7]

4800² = (6.67*10^-11 * m•) / 6*10^7

2.3*10^7 * 6*10^7 = 6.67*10^-11 * m•

1.38*10^15 = 6.67*10^-11 * m•

m• = 1.38*10^15 / 6.67*10^-11

m• =2.07*10^25 kg

Having found that, we use the value to find our vs2

vs2 = √[(G.m•) / r2]

vs2 = √[(6.67*10^-11 * 2.07*10^25) / 3*10^7]

vs2 = √(1.38*10^15 / 3*10^7)

vs2 = √4.6*10^7

vs2 = 6782.33 m/s

Therefore, the orbital speed of the second satellite is 6782 m/s