Why do we not feel weightless at equator but feel in satellite

A person living in his house at the equator goes in a circular orbit of radius equal to the radius of the earth. My question is, why does he/she not feel weightless as a satellite passenger does?

If we compare a geostationary satellite with the earth's equatorial surface then we know they both revolve around the centre of earth with same Angular velocity. So if Normal force is zero in satellite then why not at equator's surface.

• Maybe altitude has something to do with it? – ZeroTheHero Mar 4 '17 at 15:11
• But suppose we have launched a satellite very close to earth's surface such that altitude h <<< R (radius of earth), then also we see that a person in satellite experience weightless. Why? – Avi Mar 4 '17 at 15:14
• You can feel weightless at the equator. For example if you go skydiving or bungee jumping or take a very fast down elevator. Now, what do those things have in common? – The Photon Mar 5 '17 at 0:48

Hint:

The reason for thinking that you are weightless when in an orbiting satellite is that the satellite and your self would be accelerating towards the Earth at exactly the same rate. So there is no normal reaction between you and the satellite.

You must have a force acting on you because otherwise you would not accelerate and hence not be in orbit!

Draw a free body diagram of a satellite/space shuttle in orbut and an astronaut inside and you'll get the picture better.

This is tagged as a homework question so I'm not going to answer it outright, only point you to what might help

A person living in his house at the equator goes in a circular orbit of radius equal to the radius of the earth. My question is, why does he/she not feel weightless as a satellite passenger does?

What are the forces involved and what are their equations? (answer behind the spoiler but try yourself first)

Centripetal force and Gravitational given by $F_{c}= m \omega^{2} r$ and $F_{g}=\frac{GmM}{r^{2}}$ Where the net force (towards Earth) is given by $F_{g}-F_{c}$

And now consider what is changing between your two positions (Earth and orbit) and what our net force would be according to this change. Remember your point about the angular velocity ($\omega$) being the same for both cases and think about what else may be constant.

• I known that equation at earth's equatorial surface will be $N+m\omega^{2} r=mg$. And the equation is same for the satellite also but I won't understand why we take there $N=0$ as we don't do it on earth's surface. – Avi Mar 4 '17 at 15:40
• Your equation mg isn't quite correct for the equivalent in space. If you hover over the second box in my answer you'll see the equation you should be using. – Lio Elbammalf Mar 4 '17 at 16:32
• @Lio Elbammalf, the "hidden answer" is a clever trick. – David White Mar 4 '17 at 21:00

The fact that there is a special orbit called "geostationary" should ring a bell. What is so special about this orbit, corresponding to a specific radius? It is the only radius at which the satellite's angular speed is the same as the angular speed of a point on the surface of the Earth (the point has to be on the Equator). For all the other orbits the orbital speed will result in different angular speeds. All orbits below this have faster angular speeds. All the orbits above the geostationary have slower angular speeds.

For example, for a satellite orbiting just a little bit above the Equator, the orbital speed is about 8 km/s whereas the points on the Equator move with a little less than 500 m/s. So there is a factor of about 16 between the two speeds which corresponds to a factor of about 256 between accelerations (centripetal acceleration is proportional to speed squared). This explain why moving with the Earth you don't feel weightless. At the speed of the satellite, the centripetal acceleration is equal to the gravitational acceleration and you feel weightless. If you move with the Earth, your centripetal acceleration is about 250 times less than the one of the satellite corresponding to the very low orbit. The effects is a reduction of gravitational acceleration by a factor of about 260 or about 10/260 or about 0.04 $m/s^2$

You do weigh slightly less at the equator than on the poles. however escape velocity is much greater than rotational velocity.

Note: This answer contains the complete solution. The answer has been written such that the more you read, the more clues you get. At the end, you'll find the complete answer. Every line is a new hint. Please read each line and try to solve the question yourself. If you fail to solve, read the next line and try again.

The solution is in the following formula:

$$F = \frac{mv^2}{r}$$

The earth takes 24 hours to rotate once. From that information, you can calculate your tangential velocity for a person on the surface to be around $500ms^{-1}$.

The centripetal force you need for $500ms^{-1}$ is:

$$F = \frac{mv^2}{R} = m0.04\space N$$

The earth excerts a force of $mg$ that is nearly equal to $10m$. This value is much bigger than the required centripetal force. Hence, you feel weight.

The people in the sattelites are orbiting at a velocity such that the entire gravitational force is utilized as centripetal force.

$$mg_\text{at that point} = \frac{mv^2}{r}$$

$$v_{orbital} = \sqrt{\frac{GM}{r}}$$

For the specific case of geostationary satellites, their orbit is $36,000,000m$ away. The gravitational force due to the earth is quite small and this is nearly equal to the centripetal force required at that orbit. Therefore, the people there feel weightless.

• So do you mean that in geostationary satellites we ignore Normal weight as it is too small. – Avi Mar 4 '17 at 15:43
• You read the whole answer so soon? I am not going to clarify any more. The answer has everything. You need to think and solve it yourself. – Yashas Mar 4 '17 at 15:44
• I wrote the answer such that each line was a hint to solve the problem and I wanted you to try to solve it yourself after reading each line. I won't trust OPs next time. – Yashas Mar 4 '17 at 15:46
• Yes you were right, I readed question too fast without solving for each line. Sorry for that but now I am reading it again. I will ask you again if any doubt arises. – Avi Mar 4 '17 at 15:49
• Please do not post complete solutions to homework-like questions. Our policy on this can be found here which includes: “If someone posts an answer to a homework-type question that gives away a complete or near-complete solution, in most cases it will be temporarily deleted.” Please consider deleting this answer yourself. – garyp Mar 5 '17 at 3:30

protected by Qmechanic♦Mar 4 '17 at 18:17

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