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alex41 [277]
3 years ago
11

A 320-kg satellite experiences a gravitational force of 800 N. What is the radius of the satellite’s orbit? What is its altitude

?
Physics
1 answer:
Westkost [7]3 years ago
3 0

Answer:

Radius of orbit = 3.992 × 10^{7} m

Altitude of Satellite =33541.9×  m

Explanation:

Formula for gravitational force for a satellite of mass m  moving in an orbit of radius r around a planet of mass M is given by;

F= G\frac{m M}{r^{2} }

Where G = Gravitational constant = 6.67408 × 10-11 \frac{m^{3} }{Kg sec^{2} }

We are given

F= 800 N

m = 320 Kg

M = 5.972 × 10^{24} Kg

G = 6.67408 × 10-11 \frac{m^{3} }{Kg sec^{2} }

We have to find radius r =?

putting values in formula;

==> 800 =6.67408 ×10^{-11}× 320 × 5.972 × 10^{24} / r^{2}

==> 800= 39.8576 × 10^{13} × 320 / r^{2}

==> 800 = 12754.43 × 10^{13} / r^{2}

==> r^{2} = 12754.43 × 10^{13} /800

==> r^{2}=15.94 × 10^{13}

==> r =  3.992 × 10^{7} m

==> r = 39920×10^{3} m

This is the distance of satellite from center of earth. To find altitude we need distance from surface of earth. So we will subtract radius of earth from this number to find altitude.

Radius of earth =6378.1 km = 6378.1 × 10^{3} m

Altitude = 39920×10^{3} - 6378.1 ×

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A long, straight wire of radius R carries a steady current I that is uniformly distributed through the cross section of the wire
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From the question we are told that

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When r \ge R

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4 0
4 years ago
Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are
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Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

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(d₁₂)² = 18 m²

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(d₁₃)² = 10 m²

d_{14} =\sqrt{3^{2}+2^{2}  } = \sqrt{13} m  : distance from q₁ to q₄

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q₁=  7*10⁻⁹C

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α= tan⁻¹(3/3)= 45°  ,  β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

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Fn₁y = F₂₁y+ F₃₁y+F₄₁y

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Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N  

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F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }

F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }

Fn₁= 23.95*10⁻⁹ N

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