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bearhunter [10]
3 years ago
15

when a satellite is a distance d from the center of the earth, the force due to gravity on the satellite is F. what would be the

force due to gravity on the satellite when it's distance from the center of the earth is 3d?
Physics
1 answer:
swat323 years ago
5 0
F₁ = c / d²
F₂ = c / (3d)²

F₁/F₂ = 3² = 9

F₂ = 1/9 F₁

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Read 2 more answers
If position vector r = bt^2i + ct^3j, where b and c are positive constants, when does the velocity vector make an angle of 450 w
vazorg [7]

\vec r(t)=bt^2\,\vec\imath+ct^3\,\vec\jmath

The velocity at time t is

\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=2bt\,\vec\imath+3ct^2\,\vec\jmath

Take two vectors that point in the positive x and positive y directions, such as \vec\imath and \vec\jmath. The dot products of the velocity vector with \vec\imath and \vec\jmath are

\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\cdot\vec\imath=2bt=\sqrt{4b^2t^2+9c^2t^4}\cos\theta

and

\dfrac{\mathrm d\vec r(t)}{\mathrm dt}\cdot\vec\jmath=3ct^2=\sqrt{4b^2t^2+9c^2t^4}\cos\theta

We want the angles between these vectors to be 45º, for which we have \cos45^\circ=\frac1{\sqrt2}. So

\begin{cases}2\sqrt2\,bt=\sqrt{4b^2t^2+9c^2t^4}\\3\sqrt2\,ct^2=\sqrt{4b^2t^2+9c^2t^4}\end{cases}\implies3\sqrt2\,ct^2-2\sqrt2\,bt=0

\implies t(3ct-2b)=0

\implies t=0\text{ or }t=\dfrac{2b}{3c}

When t=0, the velocity vector is equal to the zero vector, which technically has no direction/doesn't make an angle with any other vector. So the only time this happens is for

\boxed{t=\dfrac{2b}{3c}}

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