Question regarding gravity and time I and my friend (Age 15) were discussing about light and speed of light when we thought of a question. Imagine you are travelling in a spaceship at the speed of 2.9*10^8 m/s circling the earth. According to our calculations, 1 second on the spaceship is 30 seconds on earth. Lets take the gravity on earth to be 10m/s^2. Now, what will be the force of gravity acting on a person in the spaceship? We tried to find concepts regarding this but could not find anything.
 A: Start in the Earth frame i.e. the frame of an observer standing on the Earth watching the orbiting spaceship. In this frame the acceleration is given by the usual expression for circular motion:
$$ a = \frac{v^2}{r} $$
where $v$ is the speed of the spaceship and $r$ is its distance from the centre of the Earth.
This is less than the acceleration measured on board the spaceship because on the spaceship time is dilated. Suppose someone on the spaceship drops an object and we on Earth watch it for a time $dt$, then distance we see the object move is given by the standard Newtonian expression:
$$ dr = \tfrac{1}{2}a\,dt^2 \tag{1} $$
The observers on the spaceship also see the object fall a distance $dr$ because radial distance is the same in both frames. However time is dilated on the spaceship because of the motion, so the time the object takes to fall is $dt'$ where:
$$ dt' = \frac{dt}{\gamma} $$
So on the spaceship the distance the object moves is given by:
$$ dr = \tfrac{1}{2}a'\,dt'^2 = \tfrac{1}{2}a'\,\frac{dt^2}{\gamma^2} \tag{2} $$
whre $a'$ is the accelkeration on the spaceship. Since it's the same $dr$ in both equations (1) and (2) we can equate the right hand sides of the two equations to get:
$$ \tfrac{1}{2}a\,dt^2 = \tfrac{1}{2}a'\,\frac{dt^2}{\gamma^2} $$
And this gives us:
$$ a' = \gamma^2 a = \gamma^2 \frac{v^2}{r} $$
To calculate a value for $a'$ you need to specify the distance $r$ at which the spaceship is circling, and you haven't specified this in your question.
One last note, the Lorentz factor $\gamma$ is given by:
$$ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} $$
At $v = 2.9 \times 10^8$ m/sec I get $\gamma \approx 3.91$ so one second on the ship is 3.91 seconds on Earth. I don't understand where you get the factor of 30 from.
A: "Imagine you are travelling in a spaceship at the speed of 2.9*10^8 m/s circling the earth"
Your speed of .97c is too large to be in a circular orbit around the earth unless all the mass of the earth were compressed to almost its Schwarzschild radius of $r_s=.9  cm$.  The circular orbit speed about this ball would be c at a radius of $1.5r_s=1.8 cm$ (https://en.wikipedia.org/wiki/Schwarzschild_radius).  Your orbit for .97c would be at a slightly larger radius.
"Now, what will be the force of gravity acting on a person in the spaceship?"
Let's continue with the picture of your now tiny space ship in orbit about what would be black-hole earth.  If the person in the spaceship were tiny enough, he would feel no acceleration (nor would the walls of the spaceship push on him) because he and the spaceship are in freefall !!  This is just like astronauts floating around in the Space Station in orbit around the earth.  If he was of finite size and his feet were pointing at earth, then his feet would accelerate away from his head (he would feel stretched), and he would feel himself compressed sideways. This happens to astronauts in the space station too, but it is way too small for them to feel. This is rightly known as tidal acceleration since it is the difference of the moon's acceleration on the near and far sides of the earth that cause the ocean tides.  
We can also ask what a free falling observer at some other R (for instance far away at large radius near infinity along a radial line or near the black-hole surface at small r) would see for the acceleration of your astronaut.  This observer would be using his own rulers and clock.
The free falling observer at infinity (and at rest wrt astronaut), using his own rods and clock, will measure the standard coordinate acceleration of the astronaut and spaceship.
$$
a(R=\infty)=\frac{GM_{earth}}{r^2}=\frac{v^2}{r}
$$
The free falling observer right next to the astronaut (and at rest wrt astronaut) at r will measure the astronaut's acceleration to be
$$
a(R=r)=0
$$
I have to think about what the general expression a(R) is, including both the effect of gravity and the effect of the relative velocity .97c between the observer and the astronaut.  This may not be simple.
