Question

A rocket is shot straight up from the earth, with a net acceleration (= acceleration by the rocket engine minus gravitational pullback) of during the initial stage of flight until the engine cut out at sec. How high will it go, air resistance neglected?

Answer 1

The acceleration of a rocket is given by

$$\dfrac{d^2s}{dt^2}=7t$$

The initial velocity of rocket at the time $t=0$  is zero

$$v(0)=0$$

Consider

$$v(t)=\dfrac{ds}{dt}$$

Then

$$\dfrac{dv}{dt}=7t$$

$$v=\int7tdt$$

$$v(t)=\dfrac{7}{2}t^2+v(0)$$

$$v(t)=\dfrac{7}{2}t^2$$

$$s(t)=\int\dfrac{7}{2}t^2dt$$

$$s(t)=\dfrac{7}{6}t^3+s(0)$$

At $t = 0,$ the rocket is on the surface of the earth, then

$$s(t)=\dfrac{7}{6}t^3+R_E$$

The engine cuts out at $t = 10 s$ . The velocity at engine cutoff is

$$v_1=v(10)=\dfrac{700}{2}\ m/s=350\ m/s$$

The distance of the rocket from the center of the earth at engine cutoff is

$$s_1=s(10)=\dfrac{7000}{6}\ m+R_E$$

$$\approx6379267\ m$$

In the unpowered phase of flight, only gravity acts on the rocket. So we need to figure out how far up an object that starts at $s=s_1$ with velocity $v = v_1$ will go.

We have

$$v\dfrac{dv}{ds}=-\dfrac{gR_E^2}{s^2}$$

$$vdv=-\dfrac{gR_E^2}{s^2}ds$$

Integrate

$$\dfrac{v^2}{2}=\dfrac{gR_E^2}{s}+\dfrac{C}{2}$$

$$v^2=\dfrac{2gR_E^2}{s}+C$$

$$C=v_1^2-\dfrac{2gR_E^2}{s_1}$$

The rocket will be at it’s highest point (greatest value of s) when it comes to a stop, just before starting to fall back to earth

$$v=0=>\dfrac{2gR_E^2}{s}+v_1^2-\dfrac{2gR_E^2}{s_1}=0$$

$$s=\dfrac{2gR_E^2s_1}{2gR_E^2-v_1^2s_1}$$

$$s\approx\dfrac{2(9.81\ m/s^2)(6378100\ m)^2(6379267\ m)}{2(9.81\ m/s^2)(6378100\ m)^2-(350\ m/s)^2(6379267\ m)}$$

$$\approx6385519\ m$$

Thus, the rocket reachs about

$$6385519\ m-6378100\ m=7419\ m$$

above the surface of the earth.