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Assoli18 [71]
3 years ago
8

A satellite is orbiting the earth. If a payload of material is added until it doubles the satellite's mass, the earth's pull of

gravity on this satellite will double but the satellite's orbit will not be affected. A satellite is orbiting the earth. If a payload of material is added until it doubles the satellite's mass, the earth's pull of gravity on this satellite will double but the satellite's orbit will not be affected. True False
Physics
2 answers:
adelina 88 [10]3 years ago
7 0

Answer:

it is True

Explanation:

I hope this helps

Fofino [41]3 years ago
5 0

Answer:

Explanation:

True, provided that the extra mass is also already in the same orbit itself.

If the extra mass is NOT in the same orbit, it's called a collision and the orbits of both will be altered due to the changes in momentum and kinetic energy.

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Seema knows the mass of a basketball. What other information is needed to find the ball’s potential energy?
adelina 88 [10]

Potential energy can be found using this formula:

PE= m * g * h

where:

PE= potential energy

m=mass

g=gravitational acceleration constant (9.8 m/s^2)

h= height

So your answer is height because you also use the gravitational constant.

7 0
3 years ago
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
3 years ago
Cevab 2 anlatarak çözer misiniz​
Assoli18 [71]
Translate please, i’d be able to help better:)
6 0
3 years ago
A total positive charge of 12.00 mC is evenly distributed on a straight thin rod of length 6.00 cm.
Gala2k [10]
It’s not in English can you change it please
3 0
3 years ago
In the diagram of the earth’s interior, which part causes the diffraction of P waves made by earthquakes?
damaskus [11]

Answer:

D

Explanation:

4 0
3 years ago
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